Solution: Option(C) To find: Area of $A B C$ Given: $A(3,-2), B(k, 2)$ and $C(8,8)$ The formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} &...
Mark the tick against the correct answer in the following: If the points
Mark the tick against the correct answer in the following: The solution set of the equation
is
A.
B.
C.
D. None of these
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8\end{array}\right|=0$ Applying $\mathrm{R}_{1}...
Mark the tick against the correct answer in the following: The solution set of the equation
is
A.
B.
C.
D. None of these
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0$ Applying...
Mark the tick against the correct answer in the following: The solution set of the equation
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \{4,8\}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-513fda79e99ca0a8956db1a94ae612f9_l3.png)
Solution: Option(A) To find: Value of $x$ We have, $\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ x-4 & 2 x-9 & 3 x-16 \\ x-8 & 2 x-27 & 3 x-64\end{array}\right|=0$ Applying...
Mark the tick against the correct answer in the following: If
then
A. 0
B. 6
C.
D. 9
Solution: Option(C) To find: Value of $\mathrm{x}$ We have, $\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & 2 \\ 9 & 6 & -2\end{array}\right|=0$ Applying $R_{1} \rightarrow 2...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com a^{2} b^{2} c^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4ffedc20a067070c39b0031dbd9ef5a1_l3.png)
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}a & 1 & b+c \\ b & 1 & c+a \\ c & 1 & a+b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & 1 &...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. None of these
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{a}+\mathrm{c} & 1 \\ \mathrm{ab} &...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com (a+b+c)(b-c)^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6b8f77728a197b0167a3206e407d475a_l3.png)
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}b+c & a & b \\ c+a & c & a \\ a+b & b & c\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following:
A.
B.
C. 0
D. none of these
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|$ We...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following: If
be distinct positive real numbers then the value of
is
A. positive
B. negative
C. a perfect square
D. 0
Solution: Option(B) To find: Nature of $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & b & c \\...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com a b c(a-b)(b-c)(c-a)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f18e308bdb4d7f9e1e29f013113bdb90_l3.png)
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1...
Mark the tick against the correct answer in the following:
A. 0
B. 1
C.
D. none of these
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 1+3 p+2 q \\ 3 & 6+3 p & 1+6 p+3 q\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. 0
Solution: Option(D) To find: Value of $\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ We have, $\left|\begin{array}{lll}a-b...
Mark the tick against the correct answer in the following:
A. 8
B.
C. 16
D. 142
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2}\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following: If
is a complex root of unity then
A. 1
B.
C. 0
D. none of these
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|$ We have,...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \cos 16^{\circ}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7221d7d9b9708c863c0710e461434ef2_l3.png)
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\sin 23^{\circ} & -\sin 7^{\circ} \\ \cos 23^{\circ} & \cos 7^{\circ}\end{array}\right|$ Formula used: (i) $\sin (A+B)=\sin A...
Mark the tick against the correct answer in the following:
A. 1
B. 0
C.
8D. ![Rendered by QuickLaTeX.com \sin 50^{\circ}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9ac1f1be82e047ba752fc04c3f57df5c_l3.png)
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\cos 70^{\circ} & \sin 20^{\circ} \\ \sin 70^{\circ} & \cos 20^{\circ}\end{array}\right|$ Formula used: (i) $\cos \theta=\sin...
If the points
and
are collinear, prove that ![Rendered by QuickLaTeX.com \frac{1}{a}+\frac{1}{b}=1](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ed373150969f07fb1661ac8bf7b6f32a_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
If
and
be three points such that area of a
is 4 sq units, find the value of
.
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the value of
for which the area of a ABC having vertices
and
is 35 sq units.
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the value of
for which the points
and
are collinear.
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the value of
for which the points
and
are collinear.
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Use determinants to show that the following points are collinear.
and ![Rendered by QuickLaTeX.com R(-5,-4)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fd5c78358e54f7c2902030112f53c513_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Use determinants to show that the following points are collinear.
and ![Rendered by QuickLaTeX.com C(10,14)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5deb6e4c4e20f7e78f807dbbc34aabf2_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Use determinants to show that the following points are collinear.
and ![Rendered by QuickLaTeX.com C(5,8)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cc1ec3609415d61387386f89725dd44a_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the area of the triangle whose vertices are:
and ![Rendered by QuickLaTeX.com R(10,8)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1aaf401ff2b3d4bb3b5d0a8f2ef9798b_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the area of the triangle whose vertices are:
and ![Rendered by QuickLaTeX.com R(4,3)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c4c4861ffeb040f077ac94b42a82d08_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the area of the triangle whose vertices are:
and ![Rendered by QuickLaTeX.com C(-1,5)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-add32e746e0d76d57e8a4d88a682c9d2_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Find the area of the triangle whose vertices are:
and ![Rendered by QuickLaTeX.com C(5,-1)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-09023f26dd604aea3dbcea1fefb32ccd_l3.png)
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Prove that ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-086096ead65e67a163434d46d12fd028_l3.png)
Solution: Operating $\mathrm{C}_{1} \rightarrow \mathrm{aCl}_{1}$ $=\frac{1}{a}\left|\begin{array}{ccc} a^{2} & b-c & c+b \\ a^{2}+a c & b & c-a \\ a^{2}-a b & b+a & c...
Solve the following equations: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} \mathrm{x} & 3 & 7 \\ 2 & \mathrm{x} & 2 \\ 7 & 6 & \mathrm{x} \end{array}\right|=0](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f1883308830b4f39885a4fff2f0a43d3_l3.png)
Solution: Operating $R 1 \rightarrow R 1+R_{2}+R_{3}$ $0=\left|\begin{array}{ccc} x+9 & x+9 & x+9 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right|$ Taking $(x+9)$ common from...
Solve the following equations: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array}\right|=0](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-19c97800f2d003d6380d2bacc131c85f_l3.png)
Solution: Operating $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ $0=\left|\begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array}\right|$ Taking $(x+9)$ common...
Solve the following equations: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} 3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8 \end{array}\right|=0](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1a5c06d09d9bf7a0d02478b25b037e49_l3.png)
Solution: $\text { Operating } \mathrm{Cl} \rightarrow \mathrm{Cl}+\mathrm{C}_{2}+\mathrm{C}_{3}$ $0=\left|\begin{array}{ccc} 3 x-8+3+3 & 3 & 3 \\ 3+3 x-8+3 & 3 x-8 & 3 \\ 3+3+3 x-8...
Solve the following equations: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & x^{3} \end{array}\right|=0](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd0959f1b595d1c650196a1bed2f7dc6_l3.png)
Solution: Operating $R 1 \rightarrow R 1^{-} R_{2}, R_{2 \rightarrow R 2}-R_{3}$ $\left|\begin{array}{ccc} 0 & x-b & x^{3}-b^{3} \\ 0 & b-c & b^{3}-c^{3} \\ 1 & c & c^{3}...
Show that
is a root of the equation
.
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}$ $\begin{aligned} 0 &=\left|\begin{array}{ccc} x-2 & -6+3 x & -1-x+3 \\ 2 & -3 x & x-3 \\ -3 & 2 x & 2+x...
Without expanding the determinant, prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|=\left|\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-79ee66c4b51421b68528b97b79ec8a52_l3.png)
Solution: $\text { Operating } R 1 \rightarrow R 1^{-} R_{3}, R_{2 \rightarrow R_{2}-R 3}$ $\begin{array}{l} \left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{c} &...
Without expanding the determinant, prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-168c8576895edb7aaa222c31079cdf05_l3.png)
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\rightarrow\left|\begin{array}{ccc} 0 & \mathrm{a}-\mathrm{b} & \mathrm{b} c-\mathrm{ac} \\ 0 &...
Prove that ![Rendered by QuickLaTeX.com \left|\begin{array}{lll}1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3}\end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32ed57c58da6e403f4ee9fa4c90e0314_l3.png)
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{array}{l} =\left|\begin{array}{ccc} 0 & \mathrm{a}^{2}+\mathrm{bc}-\mathrm{b}^{2}-\mathrm{ac} &...
If
and
, prove that ![Rendered by QuickLaTeX.com x y z(x y+y z+z x)=(x+y+z)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17d21315a42c395ba21050acc8ef3843_l3.png)
Solution: By properties of determinants, we can split the given determinant into 2 parts $\rightarrow 0=\left|\begin{array}{lll} \mathrm{x} & \mathrm{x}^{3} & \mathrm{x}^{4} \\ \mathrm{y}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-79edddeca4c6ab31f9d6be1088a59f01_l3.png)
Solution: Operating $R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$ $\begin{array}{l} =\left|\begin{array}{ccc} (a+1)(a+2)-(a+2)(a+3) & a+2-a-3 & 0 \\ (a+2)(a+3)-(a+3)(a+4)...
Using properties of determinants prove that:
, where
are in AP.
Solution: Given that $\alpha, \beta, \gamma$ are in an $A P$, which means $2 \beta=\alpha+\gamma$ Operating $R_{3} \rightarrow R_{3}-2 R_{2}+R_{1}$ $\begin{array}{l} =\left|\begin{array}{ccc} x-3...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} -\mathrm{a}\left(\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{a}^{2}\right) & 2 \mathrm{~b}^{3} & 2 \mathrm{c}^{3} \\ 2 \mathrm{a}^{3} & -\mathrm{b}\left(\mathrm{c}^{2}+\mathrm{a}^{2}-\mathrm{b}^{2}\right) & 2 \mathrm{c}^{3} \\ 2 \mathrm{a}^{3} & \mathrm{ab}^{3} & -\mathrm{c}\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right) \end{array}\right|=(\mathrm{abc})\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)^{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d60d87bff0c56f6fbf24914aa937ff27_l3.png)
Solution: Taking $a, b, c$ from $C_{1}, C_{2}, C_{3}$ $=a b c\left|\begin{array}{ccc} -b^{2}-c^{2}+a^{2} & 2 b^{2} & 2 c^{2} \\ 2 a^{2} & b^{2}-c^{2}-a^{2} & 2 c^{2} \\ 2 a^{2} &...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} (b+c)^{2} & a b & c a \\ a b & (a+c)^{2} & b c \\ a c & b c & (a+b)^{2} \end{array}\right|=2 a b c(a+b+c)^{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8c9143ae611c7adcef159f2cbce3bbfd_l3.png)
Solution: $=\left|\begin{array}{ccc} b^{2}+c^{2}+2 b c & a b & a c \\ a b & a^{2}+c^{2}+2 a c & b c \\ a c & b c & a^{2}+b^{2}+2 a b \end{array}\right|$ Operating $R_{1}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} a & b-c & c+b \\ a+c & b & c-a \\ a-b & a+b & c \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9b3c6f1a6a81774d524152d9173b8dc6_l3.png)
Solution: Operating $\mathrm{C}_{1} \rightarrow \mathrm{aCl}_{1}$ $\frac{1}{a}\left|\begin{array}{ccc} a^{2} & b-c & c+b \\ a^{2}+a c & b & c-a \\ a^{2}-a b & a+b & c...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-902b58bdec6890922608807596c3c896_l3.png)
Solution: Operating $R_{1} \rightarrow R_{1}+b R_{3}, R_{2} \rightarrow R_{2^{-}} a R_{3}$ $\begin{array}{l} \left|\begin{array}{ccc} 1+a^{2}-b^{2}+2 b^{2} & 2 a b-2 a b & -2 b+b-a^{2}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} b^{2}+c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2}+a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2}+b^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-50c84c7eae501dfc5336a66832130464_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} b^{2}+c^{2} & a^{2} & a^{2} \\ b^{2} & c^{2}+a^{2} & b^{2} \\ c^{2} & c^{2} & a^{2}+b^{2} \end{array}\right| \\...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|=\left(a^{2}+b^{2}+c^{2}\right)(a-b)(b-c)(c-a)(a+b+c)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d3b66d23a651b2d603b5732f754584e_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \\...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \begin{array}{l} \left|\begin{array}{ccc} (\mathrm{m}+\mathrm{n})^{2} & 1^{2} & \mathrm{mn} \\ (\mathrm{n}+1)^{2} & \mathrm{~m}^{2} & \ln \\ (1+\mathrm{m})^{2} & \mathrm{n}^{2} & \operatorname{lm} \end{array}\right|=\left(1^{2}+\mathrm{m}^{2}+\mathrm{n}^{2}\right)(1-\mathrm{m}) \\ (\mathrm{m}-\mathrm{n})(\mathrm{n}-1) \end{array}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dfb168f7c8358faedd5011bd607ea43c_l3.png)
Solution: $\left|\begin{array}{ccc}(\mathrm{m}+\mathrm{n})^{2} & \mathrm{l}^{2} & \mathrm{mn} \\ (\mathrm{n}+\mathrm{l})^{2} & \mathrm{~m}^{2} & \mathrm{ln} \\ (1+\mathrm{m})^{2}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} (x-2)^{2} & (x-1)^{2} & x^{2} \\ (x-1)^{2} & x^{2} & (x+1)^{2} \\ x^{2} & (x+1)^{2} & (x+2)^{2} \end{array}\right|=-8](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c4822ebfa76c440d3734cc9089233f7a_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} (\mathrm{x}-2)^{2} & (\mathrm{x}-1)^{2} & \mathrm{x}^{2} \\ (\mathrm{x}-1)^{2} & \mathrm{x}^{2} & (\mathrm{x}+1)^{2} \\...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right|=2(a+b)(b+c)(c+a)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6fd0eb7784d89751de7ca801e7d7ec49_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right| \\ =\left|\begin{array}{ccc} a+b & a+b &...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c \end{array}\right|=-a^{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e54f41911c29a651b86df4c7334b5832_l3.png)
Solution: $\left|\begin{array}{ccc}a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|$...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|=3 a b c-a^{3}-b^{3}-c^{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a662017a9f0c6baa05ba647989a81ed5_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{lll} \mathrm{b}+\mathrm{c} & \mathrm{a}-\mathrm{b} & \mathrm{a} \\ \mathrm{c}+\mathrm{a} & \mathrm{b}-\mathrm{c} & \mathrm{b} \\...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \mathrm{x}^{2} & \mathrm{y}^{2} & \mathrm{z}^{2} \\ \mathrm{x}^{3} & \mathrm{y}^{3} & \mathrm{z}^{3} \end{array}\right|=\mathrm{xyz}(\mathrm{x}-\mathrm{y})(\mathrm{y}-\mathrm{z})(\mathrm{z}-\mathrm{x})](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f0c6a341849daa263e3549b17b33bb53_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \mathrm{x}^{2} & \mathrm{y}^{2} & \mathrm{z}^{2} \\ \mathrm{x}^{3} & \mathrm{y}^{3}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right|=3(x+y+z)(x y+y z+z x)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7722870ec57de789d6200469111bf90e_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} 3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z \end{array}\right| \\ =\left|\begin{array}{ccc} x+y+z & -x+y...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x \end{array}\right|=9 y^{2}(x+y)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-82ca831bc31e34da24cab42feaaab553_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} \mathrm{x} & \mathrm{x}+\mathrm{y} & \mathrm{x}+2 \mathrm{y} \\ \mathrm{x}+2 \mathrm{y} & \mathrm{x} & \mathrm{x}+\mathrm{y} \\...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} \mathrm{x}+\lambda & 2 \mathrm{x} & 2 \mathrm{x} \\ 2 \mathrm{x} & \mathrm{x}+\lambda & 2 \mathrm{x} \\ 2 \mathrm{x} & 2 \mathrm{x} & \mathrm{x}+\lambda \end{array}\right|=(5 \mathrm{x}+\lambda)(\lambda-\mathrm{x})^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e6306fbafb0009468264f06888c5773_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} \mathrm{x}+\lambda & 2 \mathrm{x} & 2 \mathrm{x} \\ 2 \mathrm{x} & \mathrm{x}+\lambda & 2 \mathrm{x} \\ 2 \mathrm{x} & 2...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} \mathrm{x} & \mathrm{a} & \mathrm{a} \\ \mathrm{a} & \mathrm{x} & \mathrm{a} \\ \mathrm{a} & \mathrm{a} & \mathrm{x} \end{array}\right|=(\mathrm{x}+2 \mathrm{a})(\mathrm{x}-\mathrm{a})^{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5bbccb36092b2c7dce868d4145beca9_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{lll} \mathrm{x} & \mathrm{a} & \mathrm{a} \\ \mathrm{a} & \mathrm{x} & \mathrm{a} \\ \mathrm{a} & \mathrm{a} & \mathrm{x}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} \mathrm{a}+\mathrm{x} & \mathrm{y} & \mathrm{z} \\ \mathrm{x} & \mathrm{a}+\mathrm{y} & \mathrm{z} \\ \mathrm{x} & \mathrm{y} & \mathrm{a}+\mathrm{z} \end{array}\right|=\mathrm{a}^{2}(\mathrm{a}+\mathrm{x}+\mathrm{y}+\mathrm{z})](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f05527a9a9d4822d5ceb70993b7ecc28_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} \mathrm{a}+\mathrm{x} & \mathrm{y} & \mathrm{z} \\ \mathrm{x} & \mathrm{a}+\mathrm{y} & \mathrm{z} \\ \mathrm{x} & \mathrm{y}...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 1+3 p+2 q \\ 3 & 6+3 p & 1+6 p+3 q \end{array}\right|=1](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-843fd3be34ca4bb68fda258e64751cb3_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 1+3 p+2 q \\ 3 & 6+3 p & 1+6 p+3 q \end{array}\right|\\ =\left|\begin{array}{ccc} -1 &...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} 1 & \mathrm{~b}+\mathrm{c} & \mathrm{b}^{2}+\mathrm{c}^{2} \\ 1 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2} \\ 1 & \mathrm{a}+\mathrm{b} & \mathrm{a}^{2}+\mathrm{b}^{2} \end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e5c38426e9a99ff123b15cc1c220816c_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{llll} 1 & \mathrm{~b}+\mathrm{c} & \mathrm{b}^{2}+\mathrm{c}^{2} \\ 1 & \mathrm{c}+\mathrm{a} & \mathrm{c}^{2}+\mathrm{a}^{2} \\ 1...
Using properties of determinants prove that: ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ b c & c a & a b \end{array}\right|=(a-b)(b-c)(c-a)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6352ade55c99cfd66c8370a198105508_l3.png)
Solution: $\begin{array}{l} \left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ b c & c a & a b \end{array}\right| \\ =\left|\begin{array}{ccc} 0 & 0 & 1 \\ a-b &...
Evaluate : ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} 1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 2^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2} \end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d97a6c03be7f6384f6cbdfdf7c9d5ee9_l3.png)
Solution: $\left|\begin{array}{lll} 1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2} \end{array}\right|=\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4...
Evaluate : ![Rendered by QuickLaTeX.com \left|\begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c312dad5614911faf3ccc0b311d0effa_l3.png)
Solution: $\left|\begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array}\right|=6 \times\left|\begin{array}{ccc} 17 & 18 & 6 \\ 1 & 6 & 4 \\...
Evaluate: ![Rendered by QuickLaTeX.com \left|\begin{array}{lll} 29 & 26 & 22 \\ 25 & 31 & 27 \\ 63 & 54 & 46 \end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7841236a756b610c9dfc891c10020a4c_l3.png)
Solution: $\left|\begin{array}{lll}29 & 26 & 22 \\ 25 & 31 & 27 \\ 63 & 54 & 46\end{array}\right|$ $=\left|\begin{array}{ccc}4 & -5 & -5 \\ 25 & 31 & 27 \\ 63...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\sqrt{3} & \sqrt{5} \\ -\sqrt{5} & 3 \sqrt{3}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ea5bbec2bcbd4a4da17dc153341825b0_l3.png)
Solution: $\left|\begin{array}{cc}\sqrt{3} & \sqrt{5} \\ -\sqrt{5} & 3 \sqrt{3}\end{array}\right| \cdot=3 \sqrt{3} \times \sqrt{3}-(-\sqrt{5} \times \sqrt{5})$ $=14$
For what value of
, the given matrix
is a singular matrix?
Solution: For $A$ to be singular matrix its determinant should be equal to 0 . $\begin{array}{l} 0=(3-2 x) \times 4-(x+1) \times 2 \\ 0=12-8 x-2 x-2 \\ 0=10-10 x \\ x=1 \end{array}$
Without expanding the determinant, prove that
. SINGULAR MATRIX A square matrix
is said to be singular if
. Also,
is called non singular if
.
Solution: We know that $C_{1} \Rightarrow C_{1}-C_{2}$, would not change anything for the determinant. Applying the same in above determinant, we get $\left[\begin{array}{lll}40 & 1 & 5 \\...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{lll}0 & 2 & 0 \\ 2 & 3 & 4 \\ 4 & 5 & 6\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32f51a6de1804af917fa79d89813b6ed_l3.png)
Solution: We know that expansion of determinant with respect to first row is $a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}$. $0(3 \times 6-5 \times 4)-2(2 \times 6-4 \times 4)+0(2 \times 5-4 \times 3)$...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3d8be1e9a2eae4e88c55cbf66d16d191_l3.png)
Solution: $\begin{array}{l} \cos 15^{\circ} \cos 75^{\circ}-\sin 75^{\circ} \sin 15^{\circ} \\ =\cos \left(15^{\circ}+75^{\circ}\right) \because \cos A \cos B-\sin A \sin B=\cos (A+B) \\ =\cos...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\cos 65^{\circ} & \sin 65^{\circ} \\ \sin 25^{\circ} & \cos 25^{\circ}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-31eca4dd143debf57f6eb5cd3161f7aa_l3.png)
Solution: By directly opening this determinant $\begin{array}{l} \cos 65^{\circ} \times \cos 25^{\circ}-\sin 25^{\circ} \times \sin 65^{\circ} \\ =\cos \left(65^{\circ}+25^{\circ}\right) \because...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{ll}\sin 60^{\circ} & \cos 60^{\circ} \\ -\sin 30^{\circ} & \cos 30^{\circ}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-88d0e92a3541f71a649886452cbc75a8_l3.png)
Solution: After finding determinant we will get, $\begin{array}{l} \operatorname{Sin} 60^{\circ}=\frac{\sqrt{3}}{2}=\cos 30^{\circ} \\ \operatorname{Cos} 60^{\circ}=\frac{1}{2}=\sin 30^{\circ} \\...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}2 \cos \theta & -2 \sin \theta \\ \sin \theta & \cos \theta\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fb446379f26715435ce3931fc5938c4f_l3.png)
Solution: After finding determinant we will get a trigonometric identity. $\begin{array}{l} 2 \cos ^{2} \theta+2 \sin ^{2} \theta \\ =2 \\ \because \sin ^{2} \theta+\cos ^{2} \theta=1...
Evaluate
.
Solution: Find determinant $\begin{array}{l} \sqrt{6} \times \sqrt{24-\sqrt{2}} 20 \times \sqrt{5} \\ \sqrt{1} 144-\sqrt{1} 100 \\ =12-10 \\ =2 \end{array}$
If
, find the value of
.
Solution: Find the determinant of $A$ and then multiply it by 3 $\begin{array}{l} |A|=2 \\ 3|A|=3 \times 2 \\ =6 \end{array}$
If
, write the value of
.
Solution: Simply by equating both sides we can get the value of $x$. $\begin{array}{l} 2 x^{2}+2 x-2\left(x^{2}+4 x+3\right)=-12 \\ \Rightarrow-6 x-6=-12 \\ \Rightarrow-6 x=-6 \\ \Rightarrow x=1...
If
, write the value of
.
Solution: This question is having the same logic as above. $\begin{array}{l} 2 x^{2}-40=18+14 \\ \Rightarrow 2 x^{2}=72 \\ \Rightarrow x^{2}=36 \\ \Rightarrow x=\pm 6 \end{array}$
If
, write the value of
.
Solution: Here the determinant is compared so we need to take determinant both sides then find $\mathrm{x}$. $\begin{array}{l} 12 x+14=32-42 \\ \Rightarrow 12 x=-10-14 \\ \Rightarrow 12 x=-24 \\...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\mathrm{a}+\mathrm{ib} & \mathrm{c}+\mathrm{id} \\ -\mathrm{c}+\mathrm{id} & \mathrm{a}-\mathrm{ib}\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-afe300cc28b1c8fe93675d27d2938b53_l3.png)
Solution: This we can very simply go through directly. $\begin{array}{l} ((a+i b)(a-i b))-((-c+i d)(c+i d)) \\ \Rightarrow\left(a^{2}+b^{2}\right)-\left(-c^{2}-d^{2}\right) \\ \Rightarrow...
Evaluate ![Rendered by QuickLaTeX.com \left|\begin{array}{cc}\mathrm{x}^{2}-\mathrm{x}+1 & \mathrm{x}-1 \\ \mathrm{x}+1 & \mathrm{x}+1\end{array}\right|](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a0399aa0c317584f6b5333ed893b4eff_l3.png)
Solution: Theorem: This evaluation can be done in two different ways either by taking out the common things anc then calculating the determinants or simply take determinant. I will prefer first...
Let
be a square matrix of order 3, write the value of
, where
.
Solution: Theorem: If $A$ be $k \times k$ matrix then $|p A|=p^{k}|A|$. Given: $p=2, k=3$ and $|A|=4$ $\begin{array}{l} |2 A|=2^{3} \times|A| \\ =8 \times 4 \\ =32 \end{array}$
If
is a
matrix such that
and
, write the value of
.
Solution: Theorem: If $A$ be $k \times k$ matrix then $|p A|=p^{k}|A|$. Given, $\mathrm{p}=4, \mathrm{k}=2$ and $|\mathrm{A}|=5$. $\begin{array}{l} |4 \mathrm{~A}|=4^{2} \times 5 \\ =16 \times 5 \\...
Mark the tick against the correct answer in the following: Domain of
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com R-\{-1,1\}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e42b56c5f4b8457babc4c40a89446155_l3.png)
Solution: Option(C) is correct. To Find: The Domain of $\sec ^{-1}(x)$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\sec ^{-1}(x)$ can be obtained from the graph...
Mark the tick against the correct answer in the following: Domain of
is
A.
B.
C.
D. None of these
Solution: Option(B) is correct. To Find: The Domain of $\cos ^{-1}(x)$ Here,the inverse function of $\cos$ is given by $y=f^{-1}(x)$ The graph of the function $y=\cos ^{-1}(x)$ can be obtained from...
Mark the tick against the correct answer in the following: Range of
is
A.
B.
C.
D. None of these
Solution: Option(C) is correct. To Find: The range of $\operatorname{cosec}^{-1}(x)$ Here,the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function...
Mark the tick against the correct answer in the following: Range of
is
A.
B.
C.
D. None of these
Solution: Option(C) is correct. To Find:The range of $\sec ^{-1}(x)$ Here,the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\sec ^{-1}(x)$ can be obtained from the graph of...
Mark the tick against the correct answer in the following: Range of
is
A.
B.
C.
D. None of these
Solution: Option(B) is correct. To Find: The range of $\tan ^{-1} x$ Here, the inverse function is given by $y=f^{-1}(x)$ The graph of the function $y=\tan ^{-1}(x)$ can be obtained from the graph...
Mark the tick against the correct answer in the following: Range of
is
A.
B.
C.
D. None of these
Solution: Option(A) is correct. To Find: The range of $\cos ^{-1} x$ Here, the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function $y=\cos ^{-1}(x)$ can be...
Mark the tick against the correct answer in the following: Range of
is
A.
B.
C.
D. None of these
Solution: Option() is correct. To Find: The range of $\sin ^{-1} x$ Here,the inverse function is given by $\mathrm{y}=\mathrm{f}^{-1}(x)$ The graph of the function $y=\sin ^{-1}(x)$ can be obtained...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{3 \pi}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ceeea125c1422911c6e0f251935d3dfa_l3.png)
Solution: Option(B) is correct. To Find: The value of $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ Now $\cot ^{-1} 9+\operatorname{cosec}^{-1} \frac{\sqrt{41}}{4}$ can be written in...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{23}{6}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-302dbe5d6cb987fb46988f397b195fb2_l3.png)
Solution: Option(B) is correct. To Find: The value of $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}$ Let $x=\cos ^{-1} \frac{4}{5}$ $\Rightarrow \cos x=\frac{4}{5}=\frac{\text {...
Mark the tick against the correct answer in the following: If
then
A.
B.
C. 3
D. 5
Solution: Option(B) is correct. Given: $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ To Find: The value of $x$ Here $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ can be written as $\tan ^{-1} x=\tan ^{-1}...
If
then
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{2 \pi}{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ffe05ecd34a92132f2ef477be0a1e077_l3.png)
Solution: Option(B) is correct. Given: $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$ To Find: The value of $\cos ^{-1} x+\cos ^{-1} y$ Since we know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$...
Mark the tick against the correct answer in the following: If
then
A. 1
B.
C. 0
D. ![Rendered by QuickLaTeX.com \frac{1}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-988f6efc6da0d36f452e3cb51635ed48_l3.png)
Solution: Option(C) is correct. To Find: The value of $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$...
Mark the tick against the correct answer in the following: If
then
A.
B.
C.
D. None of these
Solution: Option(A) is correct. To Find: The value of $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$ Now, $\tan ^{-1} x=\tan ^{-1} 1-\tan ^{-1} \frac{1}{3}\left(\because \tan...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(B) is correct. To Find: The value of $\sin \left(2 \tan ^{-1} \frac{5}{8}\right)$ Let, $x=\sin \left(2 \tan ^{-1} \frac{5}{8}\right)$ We know that $2 \tan ^{-1} x=\sin...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(A) is correct. To Find: The value of $\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$ Let, $x=\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$ $\Rightarrow \mathrm{x}=\cos \left(\tan ^{-1}...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(B) is correct. To Find: The value of $2 \tan ^{-1} \frac{1}{3}$ i.e, $\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}$ Let, $x=\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{3}$ Since we...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{2 \pi}{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ffe05ecd34a92132f2ef477be0a1e077_l3.png)
Solution: Option(B) is correct. To Find: The value of $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$ Let, $x=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$ Since we know that $\tan ^{-1} x+\tan...
Mark the tick against the correct answer in the following:
A. 1
B.
C. 0
D. none of these
Solution: Option(C) is correct. To Find: The value of $\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)$ Let, $x=\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)$ $\begin{array}{l} \Rightarrow \mathrm{x}=\cot...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(C) is correct. To Find: The value of $\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$ Let, $x=\sin ^{-1}\left(\frac{-1}{2}\right)+2 \cos...
Mark the tick against the correct answer in the following: If
then
A.
B.
C.
D. none of these
Solution: Option(B) is correct. Given: $\cot ^{-1} \frac{-1}{5}=x$ To Find: The value of $\sin \mathrm{x}$ Since, $x=\cot ^{-1} \frac{-1}{5}$ $\Rightarrow \cot \mathrm{x}=\frac{-1}{5}=\frac{\text {...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{2 \pi}{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ffe05ecd34a92132f2ef477be0a1e077_l3.png)
Solution: Option(B) is correct. To Find: The value of $\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right\}$ Let, $x=\tan ^{-1}\left\{2 \cos \left(2 \sin ^{-1}...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{2}{\sqrt{10}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-69f428c59a6a6aa1546be41f6af303f4_l3.png)
Solution: Option(C) is correct. To Find: The value of $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$ Let $x=\cos ^{-1} \frac{4}{5}$ $\Rightarrow \cos x=\frac{4}{5}$ Therefore $\sin...
Mark the tick against the correct answer in the following:
A. 1
B. 0
C.
D. none of these
Solution: Option(A) is correct. To Find: The value of of $\sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}$ Let, $x=\sin \left\{\frac{\pi}{3}-\sin...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(B) is correct. To Find: The value of $\cos \left(\tan ^{-1} \frac{3}{4}\right)$ Let $x=\tan ^{-1} \frac{3}{4}$ $\begin{array}{l} \Rightarrow \tan x=\frac{3}{4} \\ \Rightarrow \tan...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(B) is correct. To Find: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ Let, $x=\cos ^{-1} \frac{3}{5}$ $\Rightarrow \cos x=\frac{3}{5}$ Now, $\sin \left(\cos ^{-1}...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{(5+\sqrt{3})}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-169f49908d662d4a4e0c93c6ddf75b78_l3.png)
Solution: Option(A) is correct. To Find: The value of $\tan \frac{1}{2}\left(\cos ^{-1} \frac{\sqrt{5}}{3}\right)$ Let, $x=\cos ^{-1} \frac{\sqrt{5}}{3}$ $\Rightarrow \cos x=\frac{\sqrt{5}}{3}$ Now,...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{-7}{12}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-23a284377b7c8e7ec51915a03eec6888_l3.png)
Solution: Option(B) is correct. To Find: The value of $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)$ Consider, $\tan \left(2 \tan ^{-1} \frac{1}{5}-\frac{\pi}{4}\right)=\tan \left(\tan...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. none of these
Solution: Option(A) is correct. To Find: The value of $\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$ Now, let $x=\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$ $\Rightarrow...
Mark the tick against the correct answer in the following:
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \pi](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9f615e9f1df751e577153a5fc90c7227_l3.png)
Solution: Option(D) is correct. To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{2 \pi}{3}\right)\right)+\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ Here, consider $\cos...
Mark the tick against the correct answer in the following: If
then
?
A.
B. 1
C. 0
D. none of these
Solution: Option(C) is correct. Given: $x \neq 0$ To Find: The value of $\cos \left(\tan ^{-1} x+\cot ^{-1} x\right)$ Now, let $x=\cos \left(\tan ^{-1} x+\cot ^{-1} x\right)$ $\begin{array}{l}...
Mark the tick against the correct answer in the following: The value of
A. 0
B. 1
C.
D. none of these
Solution: Option(1) is correct. To Find: The value of $\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)$ Now, let $x=\sin \left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{1}{2}\right)$...
Mark the tick against the correct answer in the following:
A. 0
B.
C.
D. ![Rendered by QuickLaTeX.com \pi](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9f615e9f1df751e577153a5fc90c7227_l3.png)
Solution: Option(D) is correct. To Find: The value of $\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)$ Now, let $x=\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)$ $\begin{array}{l} \Rightarrow...
Mark the tick against the correct answer in the following: The value of
is
A.
B.
C.
D. none of these
Solution: Option(C) is correct. To Find: The value of $\tan ^{-1}\left(\tan \left(\frac{3 \pi}{4}\right)\right)$ Now, let $x=\tan ^{-1}\left(\tan \left(\frac{3 \pi}{4}\right)\right)$ $\Rightarrow...
Mark the tick against the correct answer in the following: The value of
is
A.
B.
C.
D. none of these
Solution: Option(B) is correct. To Find: The value of $\operatorname{cosec}^{-1}\left(\operatorname{cosec}\left(\frac{4 \pi}{3}\right)\right)$ Now, let...
Mark the tick against the correct answer in the following: The value of
is
A.
B.
C.
D. none of these
Solution: Option(A) is correct. To Find: The value of $\cot ^{-1}\left(\cot \left(\frac{5 \pi}{4}\right)\right)$ Now, let $x=\cot ^{-1}\left(\cot \left(\frac{5 \pi}{4}\right)\right)$ $\Rightarrow...
Mark the tick against the correct answer in the following: The value of
is A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{7 \pi}{6}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1d373f2cbf9704063c08bc4f600b83ee_l3.png)
Solution: Option(C) ids correct. To Find: The value of $\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$ Now, let $x=\cos ^{-1}\left(\cos \left(\frac{13 \pi}{6}\right)\right)$ $\Rightarrow...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{5 \pi}{6}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a7b5ea7a84027711ef7b12d492393c07_l3.png)
Solution: Option(D) is correct. To Find: The Principle value of $\cot ^{-1}(-\sqrt{3})$ Let the principle value be given by $\mathrm{x}$ Now, let $x=\cot ^{-1}(-\sqrt{3})$ $\begin{array}{l}...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{7 \pi}{6}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1d373f2cbf9704063c08bc4f600b83ee_l3.png)
Solution: Option(C) is correct. To Find: The Principle value of $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)$ Let the principle value be given by $x$ Now, let $x=\sec...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{3 \pi}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ceeea125c1422911c6e0f251935d3dfa_l3.png)
Solution: Option(D) is correct. To Find: The Principle value of $\cot ^{-1}(-1)$ Let the principle value be given by $x$ Now, let $x=\cot ^{-1}(-1)$ $\Rightarrow \cot x=-1$ $\Rightarrow \cot x=-\cot...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. none of these
Solution: Option(C) is correct. To Find: The Principle value of $\tan ^{-1}(-\sqrt{3})$ Let the principle value be given by $x$ Now, let $x=\tan ^{-1}(-\sqrt{3})$ $\begin{array}{l} \Rightarrow \tan...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{\pi}{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8d2815ec916632a87c5238696c86aaae_l3.png)
Solution: Option(B) is correct. To Find: The Principle value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ Let the principle value be given by $x$ Now, let $x=\cos ^{-1}\left(\frac{-1}{2}\right)$...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. none of these
Solution: Option(A) is correct. To Find: The Principle value of $\sin ^{-1}\left(\frac{-1}{2}\right)$ Let the principle value be given by $x$ Now, let $x=\sin ^{-1}\left(\frac{-1}{2}\right)$...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. ![Rendered by QuickLaTeX.com \frac{5 \pi}{6}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a7b5ea7a84027711ef7b12d492393c07_l3.png)
Solution: Option(B) is correct. To Find: The Principle value of $\operatorname{cosec}^{-1}(2)$ Let the principle value be given by $x$ Now, let $x=\operatorname{cosec}^{-1}(2)$ $\begin{array}{l}...
Mark the tick against the correct answer in the following: The principal value of
is
A.
B.
C.
D. none of these
Solution: Option (A) is correct. To Find:The Principle value of $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ Let the principle value be given by $x$ Now, let $x=\cos...
Write down the interval for the principal-value branch of each of the following functions and draw its graph: ![Rendered by QuickLaTeX.com \operatorname{cosec}^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-085e5991abd0df13c6c46cd03afa08e0_l3.png)
Solution: Principal value branch of $\operatorname{cosec}^{-1} x$ is $\left[-\frac{\pi}{2}, 0\right) \cup\left(0, \frac{\pi}{2}\right]$
Write down the interval for the principal-value branch of each of the following functions and draw its graph: ![Rendered by QuickLaTeX.com \sec ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-17e3a545d562c08615185a02974240ef_l3.png)
Solution: Principal value branch of $\sec ^{-1} x$ is $\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right]$
Write down the interval for the principal-value branch of each of the following functions and draw its graph: ![Rendered by QuickLaTeX.com \cot ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3894dde7b08a56927bdc6d7df7ff124c_l3.png)
Solution: Principal value branch of $\cot ^{-1} x$ is $(0, \pi)$
Write down the interval for the principal-value branch of each of the following functions and draw its graph: ![Rendered by QuickLaTeX.com \tan ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fec5b1a20e300a594ca5784aedc7ebce_l3.png)
Solution: Principal value branch of $\tan ^{-1} \times$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Write down the interval for the principal-value branch of each of the following functions and draw its graph: ![Rendered by QuickLaTeX.com \cos ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-855672fe2e3c5a5db3cf929dc0d860ea_l3.png)
Solution: Principal value branch of $\cos ^{-1} x$ is $[0, \pi]$
Write down the interval for the principal-value branch of each of the following functions and draw its graph: ![Rendered by QuickLaTeX.com \sin ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-080840b46d3db253435f41ff683a98bb_l3.png)
Solution: Principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Solve for
: ![Rendered by QuickLaTeX.com \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78e757497528c5fb5393aa2fb4322f53_l3.png)
Solution: Given: $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$ We know that $\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}$ So, $\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x$ Substituting in the...
Solve for
: ![Rendered by QuickLaTeX.com \tan ^{-1} x=\sin ^{-1} \frac{1}{\sqrt{2}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3a0b779c170db87aa78fe736d5dd9863_l3.png)
Solution: To find: value of $x$ Given: $\tan ^{-1} \mathrm{x}=\sin ^{-1} \frac{1}{\sqrt{2}}$ We know that $\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$ Therefore, $\frac{\pi}{4}=\sin ^{-1}...
Solve for
: ![Rendered by QuickLaTeX.com \cos \left(\sin ^{-1} x\right)=\frac{1}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-47098cfc81ce4e11d115a53fb0263837_l3.png)
Solution: To find: value of $x$ Given: $\cos \left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{2}$ $\text { LHS }=\cos \left(\sin ^{-1} x\right)$ $\begin{array}{l} =\cos \left(\cos...
Solve for
: ![Rendered by QuickLaTeX.com \cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e4956d75f221007a29b61da2eb3b573c_l3.png)
Solution: To find: value of $x$ Formula Used: $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$. Given: $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$ $\text { LHS }=\cos \left(2 \sin ^{-1}...
Solve for
: ![Rendered by QuickLaTeX.com \cos \left(\sin ^{-1} x\right)=\frac{1}{9}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bc72f1c0788cff6992056efa304e1eea_l3.png)
Solution: To find: value of $x$ Given: $\cos \left(\sin ^{-1} x\right)=\frac{1}{9}$ $\mathrm{LHS}=\cos \left(\sin ^{-1} \mathrm{x}\right) \ldots(1)$ Let $\sin \theta=x$ Therefore $\theta=\sin ^{-1}...
Solve for x: ![Rendered by QuickLaTeX.com \tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e5cbb83758b255209de222559a28714a_l3.png)
Solution: To find: value of $x$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where xy $<1$ Given: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$...
Solve for
: ![Rendered by QuickLaTeX.com \tan ^{-1}(x+1)+\tan ^{-1}(x-1)=\tan ^{-1} \frac{8}{31}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-abe33b243b9033d57faffc98621f18a4_l3.png)
Solution: To find: value of $x$ Formula Used: $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan ^{-1}\left(\frac{\mathrm{x}+y}{1-\mathrm{xy}}\right)$ where $\mathrm{xy}<1$ Given: $\tan...
Prove that: ![Rendered by QuickLaTeX.com 2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a97085508dbdac099081247d61b04a5c_l3.png)
Solution: To Prove: $2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$ Formula Used: 1) $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \times \sqrt{1-x^{2}}\right)$ 2) $\tan ^{-1}...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-98ebda3a34fae553c49d06d4143d0d15_l3.png)
Solution: To Prove: $\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$ Proof:...
Prove that: ![Rendered by QuickLaTeX.com \cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{27}{11}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b2dd1c76d897fd96791ec3ceee86f35a_l3.png)
Solution: To Prove: $\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{27}{11}$ Formula Used: $\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \times \sqrt{1-y^{2}}+y \times...
Prove that: ![Rendered by QuickLaTeX.com \cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13}=\sin ^{-1} \frac{56}{65}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f9672cb3e739d37a5326e5cf2e825ae9_l3.png)
Solution: To Prove: $\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13}=\sin ^{-1} \frac{56}{65}$ Formula Used: $\sin ^{-1} \mathrm{x}+\sin ^{-1} \mathrm{y}=\sin ^{-1}\left(\mathrm{x} \times...
Prove that: ![Rendered by QuickLaTeX.com \cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cec69b3d338504c28ecb3ff03a21b7e4_l3.png)
Solution: To Prove: $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$ Formula Used: $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \times...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \tan ^{1} \frac{4}{3}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-084dd6d600541d2a48d8093ed6ecd265_l3.png)
Solution: To Prove: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \tan ^{-1} \frac{4}{3} \Rightarrow 2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3}$...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5a5c79bbf01b13bd2dc16d8120b9dee6_l3.png)
Solution: To Prove: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-931d3b254a8a9dd7786c4b71edb8928c_l3.png)
Solution: To Prove: $\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi$ Formula Used: $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\pi+\tan ^{-1}\left(\frac{\mathrm{x}+\mathrm{y}}{1-\mathrm{xy}}\right)$...
Prove that: ![Rendered by QuickLaTeX.com 2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-913b28a0bacad156c41112bfbb32d181_l3.png)
Solution: To Prove: $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\frac{\pi}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3e5456e9cf8ec3d00a7c3457e2adedd4_l3.png)
Solution: To Prove: $\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\frac{\pi}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof:...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-29ae5a9226fc487921651657b1bfe8cc_l3.png)
Solution: To Prove: $\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ Proof:...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c45532ddbe1e37c33c6f0d55ef261e74_l3.png)
Solution: To Prove: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{2}{11}=\tan ^{-1} \frac{3}{4}$ Formula Used: $\tan ^{-1} \mathrm{x}+\tan ^{-1} \mathrm{y}=\tan...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\frac{x}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0fd8c58fa54a03fb41368ac0985c4718_l3.png)
Solution: To Prove: $\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)=\frac{x}{2}$ Formula Used: 1) $\sin A=2 \times \sin \frac{A}{2} \times \cos \frac{A}{2}$ 2) $1+\cos A=2 \cos ^{2} \frac{A}{2}$...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1}\left(\frac{\mathrm{x}+\sqrt{\mathrm{x}}}{1-\mathrm{x}^{3 / 2}}\right)=\tan ^{-1} \mathrm{x}+\tan ^{-1} \sqrt{\mathrm{x}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0000197c607d9b1313d1d415b52ed523_l3.png)
Solution: We know that, $\tan ^{-1}\left(\frac{A+B}{1-A B}\right)=\tan ^{-1} A+\tan ^{-1} B$ Now, taking $A=x$ and $B=\sqrt{x}$ We get, $\tan ^{-1} x+\tan ^{-1} \sqrt{x}=\tan...
Prove that: ![Rendered by QuickLaTeX.com \cot ^{-1}\left(\sqrt{1+x^{2}}-x\right)=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ca9a387013f58207d57f89bc7f58755d_l3.png)
Solution: To Prove: $\cot ^{-1}\left(\sqrt{1+x^{2}}-x\right)=\frac{\pi}{2}-\frac{1}{2} \cot ^{-1} x$ Formula Used: 1) $\tan \left(\frac{\pi}{4}+A\right)=\frac{1+\tan A}{1-\tan A}$ 2)...
Prove that: ![Rendered by QuickLaTeX.com \sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=2 \cos ^{-1} x](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2086ab7e873d85534d618ffa505033d1_l3.png)
Solution: To Prove: $\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right)=2 \cos ^{-1} x$ Formula Used: 1) $\cos 2 A=2 \cos ^{2} A-1$ 2) $\cos ^{-1} A=\sec ^{-1}\left(\frac{1}{A}\right)$ Proof:...
Prove that: ![Rendered by QuickLaTeX.com \cos ^{-1}\left(2 \mathrm{x}^{2}-1\right)=2 \cos ^{-1} \mathrm{x}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-587b9cd8cbac0958f31cc89a133bb12f_l3.png)
Solution: To Prove: $\cos ^{-1}\left(2 x^{2}-1\right)=2 \cos ^{-1} x$ Formula Used: $\cos 2 A=2 \cos ^{2} A-1$ Proof: $\text { LHS }=\cos ^{-1}\left(2 x^{2}-1\right) \ldots(1)$ Let $x=\cos A \ldots$...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} \mathrm{x}+\tan ^{-1}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)=\tan ^{-1}\left(\frac{3 \mathrm{x}-\mathrm{x}^{3}}{1-3 \mathrm{x}^{2}}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7e0455c5e014d414e6a6308eecb17214_l3.png)
Solution: To Prove: $\tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x,|x|<\frac{1}{\sqrt{3}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eabde49e0d7393423fd991b3916174cf_l3.png)
Solution: To Prove: $\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)=3 \tan ^{-1} x$ Formula Used: $\tan 3 A=\frac{3 \tan A-\tan ^{3} A}{1-3 \tan ^{2} A}$ Proof: $\text { LHS }=\tan...
Prove that: ![Rendered by QuickLaTeX.com \cos ^{-1}\left(4 x^{3}-3 x\right)=3 \cos ^{-1} x, \frac{1}{2} \leq x \leq 1](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4469256dcbb7552716e2676f9d2dda5f_l3.png)
Solution: To Prove: $\cos ^{-1}\left(4 x^{3}-3 x\right)=3 \cos ^{-1} x$ Formula Used: $\cos 3 A=4 \cos ^{3} A-3 \cos A$ Proof: $\text { LHS }=\cos ^{-1}\left(4 x^{3}-3 x\right) \ldots(1)$ Let...
Prove that: ![Rendered by QuickLaTeX.com \sin ^{-1}\left(3 x-4 x^{3}\right)=3 \sin ^{-1} x,|x| \leq \frac{1}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-95a44e293a91afd5b4a17650732e3ef7_l3.png)
Solution: To Prove: $\sin ^{-1}\left(3 x-4 x^{3}\right)=3 \sin ^{-1} x$ Formula Used: $\sin 3 \mathrm{~A}=3 \sin \mathrm{A}-4 \sin ^{3} \mathrm{~A}$ Proof: $\operatorname{LHS}=\sin ^{-1}\left(3 x-4...
Prove that: ![Rendered by QuickLaTeX.com \sin ^{-1}\left(2 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}\right)=2 \sin ^{-1} \mathrm{x},|\mathrm{x}| \leq \frac{1}{\sqrt{2}}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d4acd331c828df9bbaa100c713cd0b9a_l3.png)
Solution: To Prove: $\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$ Formula Used: $\sin 2 A=2 \times \sin A \times \cos A$ Proof: $\text { LHS }=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)...
Prove that: ![Rendered by QuickLaTeX.com \tan ^{-1} x+\cot ^{-1}(x+1)=\tan ^{-1}\left(x^{2}+x+1\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c01e9835f586356ee9d04ba6c04a4a99_l3.png)
Solution: To Prove: $\tan ^{-1} x+\cot ^{-1}(x+1)=\tan ^{-1}\left(x^{2}+x+1\right)$ Formula Used: 1) $\cot ^{-1} \mathrm{x}=\tan ^{-1} \frac{1}{\mathrm{x}}$ 2) $\tan ^{-1} \mathrm{x}+\tan ^{-1}...
Find the principal value of each of the following: ![Rendered by QuickLaTeX.com \sin ^{-1}\left\{\sin \frac{3 \pi}{5}\right\}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c91ea0e90934be98c74db31694e2d5d2_l3.png)
Solution: $\begin{array}{l} \sin ^{-1}\left\{\sin \left(\frac{3 \pi}{5}\right)\right\} \\ =\sin ^{-1}\left\{\sin \left(\pi-\frac{2 \pi}{5}\right)\right\} \end{array}$ [Formula: $\sin (\pi-x)=\sin...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dc644de21807746663921c9b85985bca_l3.png)
Solution: [Formula: $\cos ^{-1}(-x)=\pi-\cos (x)$ and $\left.\sin ^{-1}(-x)=-\sin (x)\right]$ $\tan ^{-1} 1+\left(\pi-\cos ^{-1}\left(\frac{1}{2}\right)\right)+\left(-\sin...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-922fd2965e961958108bee73c25f6850_l3.png)
Solution: Putting the values of the inverse trigonometric terms $\begin{array}{l} \frac{\pi}{3}+2 \times \frac{\pi}{6} \\ =\frac{\pi}{3}+\frac{\pi}{3} \\ =\frac{2 \pi}{3} \end{array}$
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \sin \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6b776df32cd1d202616ce5542a2e8b53_l3.png)
Solution: $\sin \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)=\sin \left(\frac{\pi}{2}\right)$ [Formula: $\left.\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2}\right]$ Putting the...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \operatorname{cosec}\left(\sin ^{-1} x+\cos ^{-1} x\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3cf6ffdd48f8f197b13b99f2cfe8ad2c_l3.png)
Solution: $\operatorname{cosec}\left(\sin ^{-1} x+\cos ^{-1} x\right)=\operatorname{cosec} \frac{\pi}{2}\left[\right.$ Formula: $\left.\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$ Putting the...
Find the principal value of each of the following: ![Rendered by QuickLaTeX.com \cot \left(\tan ^{-1} x+\cot ^{-1} x\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-47835b84771b881352643cc4bb0f683b_l3.png)
Solution: $\cot \left(\tan ^{-1} x+\cot ^{-1} x\right)=\cot \left(\frac{\pi}{2}\right)$ [Formula: $\left.\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right]$ Putting value of $\cot...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11a7418e1bfe60c71fc3dc49032d7632_l3.png)
Solution: $\begin{array}{l} \sin \left\{\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right\}\left[\text { Formula: } \sin ^{-1}(-\mathrm{x})=-\sin ^{-1} \mathrm{x}\right] \\ =\sin...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3}) 3](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b23542306926f506d6ef1a4fd0ed125f_l3.png)
Solution: $\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$ Putting the value of $\tan ^{-1} \sqrt{3}$ and using the formula $\begin{array}{l} \cot ^{-1}(-x)=\pi-\cot ^{-1} x \\...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-782a4713efc5ebd34230d6e72b3d2e56_l3.png)
Solution: $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{6}\right)\right)$ [ Formula: $\tan (\pi+x)=\tan x$, as tan is positive in the third quadrant.] $=\tan...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0b0e749b4f039b3c9e1b30935a5a300c_l3.png)
Solution: $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=\cos ^{-1}\left(\cos \left(2 \pi+\frac{\pi}{6}\right)\right)$ [ Formula: $\cos (2 \pi+x)=\cos x, \cos$ is positive in the first quadrant. ]...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d491d894f59ac26d15f15ed494ba964b_l3.png)
Solution: $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left(\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right)$ [Formula: $\cos (2 \pi-x)=\cos (x)$, as cos has a positive vaule in the fourth...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ce05e0531f1db3fb3ff1aae55387aafe_l3.png)
Solution: $\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left(\tan \left(\pi-\frac{\pi}{4}\right)\right)$ [Formula: $\tan (\pi-x)=-\tan (x)$, as tan is negative in the second quadrant. ]...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c393e49a8e47b0278f6ca5517828b31b_l3.png)
Solution: $\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)=\sin ^{-1}\left(\sin \left(\pi-\frac{\pi}{3}\right)\right)$ [ Formula: $\sin (\pi-x)=\sin x)$ $=\sin ^{-1}\left(\sin \frac{\pi}{3}\right)$ [...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \operatorname{cosec}^{-1}(2)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e006e026d284675862dc56728b345b6d_l3.png)
Solution: $\operatorname{cosec}^{-1}(2)$ Putting the value directly $=\frac{\pi}{6}$
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-328b5f2e61ba79c836f1c3288b9b0fef_l3.png)
Solution: $\sec ^{-1}\left(\frac{-2}{\sqrt{3}}\right)=\pi-\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)\left[\right.$ Formula: $\left.\sec ^{-1}(-x)=\pi-\sec ^{-1}(x)\right]$ $\begin{array}{l}...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \tan ^{-1}(-\sqrt{3})](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-625e47229e03ad7687b11a306320e966_l3.png)
Solution: $\tan ^{-1}(-\sqrt{3})=-\tan ^{-1}(\sqrt{3})$ [Formula: $\left.\tan ^{-1}(-x)=-\tan ^{-1}(x)\right]$ $=-\frac{\pi}{3}$
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \operatorname{cosec}^{-1}(-\sqrt{2})](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d97b3cdb6e1fad513145be47abf33d8d_l3.png)
Solution: $\operatorname{cosec}^{-1}(-\sqrt{2})=-\operatorname{cosec}^{-1}(\sqrt{2})$ [Formula: $\left.\operatorname{cosec}^{-1}(-x)=-\operatorname{cosec}^{-1}(x)\right]$ $=-\frac{\pi}{4}$ This can...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \sec ^{-1}(-2)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-75284f8496b6c48fb50017cb200f6615_l3.png)
Solution: $\sec ^{-1}(-2)=\pi-\sec ^{-1}(2)\left[\right.$ Formula: $\left.\sec ^{-1}(-x)=\pi-\sec ^{-1}(x)\right]$ $\begin{array}{l} =\pi-\frac{\pi}{3} \\ =\frac{2 \pi}{3} \end{array}$
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \tan ^{-1}(-1)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d9d1c931cef5a553ad82eb451434eefc_l3.png)
Solution: $\tan (-1)=-\tan (1)\left[\right.$ Formula: $\left.\tan ^{-1}(-x)=-\tan ^{-1}(x)\right]$ [ We know that $\tan \frac{\pi}{4}=1$, thus $\left.\tan ^{-1} \frac{\pi}{4}=1\right]$...
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \cos ^{-1}\left(\frac{-1}{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-509e24046c7f3adae69d07eb7ac31dc1_l3.png)
Solution: $\cos ^{-1}\left(\frac{-1}{2}\right)=\pi-\cos ^{-1}\left(\frac{1}{2}\right)$ [ Formula: $\left.\cos ^{-1}(-x)=-\cos ^{-1}(x)\right]$ $=\pi-\frac{\pi}{3}$ $=\frac{2 \pi}{3}$
Find the principal value of each of the following : ![Rendered by QuickLaTeX.com \sin ^{-1}\left(\frac{-1}{2}\right)](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a3dde0a26aee9cc7af567e8acc7620c8_l3.png)
Solution: $\sin ^{-1}\left(\frac{-1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)\left[\right.$ Formula: $\left.\sin ^{-1}(-x)=\sin ^{-1}(x)\right]$ $=-\frac{\pi}{6}$
If
and
then express
in the form
, where
and
.
Solution: $\begin{array}{l} \vec{a}=3 \hat{\imath}-\hat{\jmath} \end{array}$ $\begin{array}{l} \mathrm{I} \vec{a} \mathrm{I}=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10} \\ \hat{a}=\frac{3}{\sqrt{10}}...
If
and
are two vectors such that
then prove that vector
is perpendicular to the vector
.
Solution: $\begin{array}{l} \mathrm{I} \vec{a}+\vec{b} \mathrm{I}=\vec{a} \\ (\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a} \\ \mathrm{I} \vec{a} \mathrm{I}_{2}+\vec{a} \cdot...
If
and
are two unit vectors such that
, find
.
Solution: $\vec{a}$ and $\vec{b}$ are unit vectors $\begin{array}{l} (\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\mathrm{I}(\vec{a}+\vec{b}) \mathrm{I}_{2} \\ \mathrm{I} \vec{a}...
Three vertices of a triangle are
and
. Show that it is a right – angled triangle. Also, find its other two angles.
Solution: $\overrightarrow{O A}=-\hat{\jmath}-2 \hat{k}$ $\overrightarrow{O B}=3 \hat{\imath}+\hat{\jmath}+4 \hat{k}$ $\overrightarrow{O C}=5 \hat{\imath}+7 \hat{\jmath}+\hat{k}$ $\overrightarrow{O...
Find the value of
for which the vectors
and
are perpendicular to each other.
Solution: $\begin{array}{l} \vec{a}=2 \hat{\imath}+\lambda \hat{\jmath}+3 \hat{k} \\ \vec{b}=3 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \end{array}$ Given $\vec{a} \perp \vec{b}$ $\begin{array}{l}...
If
and
be four points, show that
is perpendicular to
.
Solution: $\begin{array}{l} \overrightarrow{O A}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \\ \overrightarrow{O B}=5 \hat{\imath}+4 \hat{\jmath}-\hat{k} \\ \overrightarrow{O C}=3 \hat{\imath}+6...
If
and the coordinates of
are
, find the coordinates of
.
Solution: $\begin{array}{l} \overrightarrow{A B}=3 \hat{\imath}-\hat{\jmath}+2 \hat{k} \\ \overrightarrow{O A}=-2 \hat{\jmath}-\hat{k} \\ \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O...
The dot products of a vector with the vector
and
are 0,5 and 8 respectively. Find the vector.
Solution: $\begin{array}{l} \vec{v}=\mathrm{v} 1 \hat{\imath}+\mathrm{v} 2 \hat{\jmath}+\mathrm{v}_{3} \hat{k} \\ \vec{a}=\hat{\imath}+3 \hat{\jmath}-2 \hat{k} \\...
If
and
are unit vectors inclined at an angle
then prove that:
i.
ii. ![Rendered by QuickLaTeX.com \tan \frac{\theta}{2}=\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aab30e11b4b3f87de445e409b0db1285_l3.png)
Solution: $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ i) $\begin{aligned} &(\hat{a}+\hat{b})(\hat{a}+\hat{b})=\mathrm{I} \hat{\mathrm{a}} \mathrm{I}_{2}+\mathrm{I}...
If
and
, find
.
Solution: $\mathrm{I} \overrightarrow{\mathrm{a}} \mathrm{I}=2$ $\mathrm{I} \vec{b} \mathrm{I}=3$ $\vec{a} \cdot \vec{b}=4$ $(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})=\mathrm{I} \vec{a}...
Find the angle between
and
, when
i.
and
ii.
and ![Rendered by QuickLaTeX.com \vec{a} \cdot \vec{b}=-1](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dd33dba27df325bb4cb9c93569fe26f1_l3.png)
Solution: i) $\mathrm{I} \vec{a} \mathrm{I}=2$ $\mid \vec{b} \mathrm{I}=1$ $\vec{a} \cdot \vec{b}=\sqrt{3}$ $\mathrm{I} \vec{a} \mathrm{I} . \mid \vec{b} \mathrm{I} \cos \theta=\sqrt{3}$ $2 \cos...
Prove that
, where
and
.
Solution: $\begin{array}{l} (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=\mathrm{I} \vec{a} \mathrm{I}_{2}+\mathrm{I} \vec{b} \mathrm{I}_{2} \\ \mathrm{I} \vec{a} \mathrm{I}_{2}-\mathrm{I} \vec{b}...
Express the vector
as sum of two vectors such that one is parallel to the vector
and the other is perpendicular to
.
Solution: $\begin{array}{l} \vec{a}=6 \hat{\imath}-3 \hat{\jmath}-6 \hat{k} \\ \vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{array}$ Let $\vec{a}$ be written as sum of $\vec{c}+\vec{d}$, where...
Find the angle between
and
, if
and
.
Solution: $\begin{array}{l} \text { Given: } \vec{a}=2 \hat{\imath}-\hat{\jmath}+3 \hat{k} \\ \vec{b}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\ \vec{a}+\vec{b}=5 \hat{\imath}+5 \hat{k} \\...
Find a vector
of magnitude
, making an angle
with
– axis,
with
– axis and an acute angle
with z – axis.
Solution: $\mathrm{I} \overrightarrow{\mathrm{a}} \mathrm{I}=5 \sqrt{2}$ Also it is given that $\alpha=\frac{\pi}{4}$ $\beta=\frac{\pi}{2}, \gamma=$ acute angle. So from this $\cos...
Show that the vector
is equally inclined to the coordinate axes.
Solution: $\begin{array}{l} \vec{a}=\cos \gamma=\frac{1}{\sqrt{3}} \\ \mathrm{I} \vec{a} \mathrm{I}=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3} \end{array}$ Since direction cosines or cosines of angle made by...
Find the angles which the vector
makes with the coordinate axes.
Solution: Direction cosines gives the angle made by the vector with $x$-axis, $y-$ axis, z-axis respectively. $\begin{array}{l} \vec{a}=3 \hat{\imath}-6 \hat{\jmath}+2 \hat{k} \\ \mathrm{I} \vec{a}...