Answer: Given, $\begin{array}{l} d = 7\\ \overline n = 3\widehat i + 5\widehat j - 6\widehat k \end{array}$ The unit vector normal to the plane: ...
Answer: Given plane, 2x + y – 2z + 5 = 0 The point is (2, 1, 0).
.Answer: Given plane, $\overrightarrow r .(2\widehat i - 2\widehat j + 4\widehat k) + 5 = 0$ The cartesian form: 2x – 2y + 4z + 5 = 0 The point is (1, 1, 2).
.Answer: Given plane, $\overrightarrow r .(2\widehat i - 5\widehat j + 3\widehat k) = 13$ The cartesian form: 2x – 5y + 3z – 13 = 0 The point is (3, 4, 5).
from the plane 
.Answer: Given plane, $\overrightarrow r .(\widehat i + \widehat j + \widehat k) + 17 = 0$ The cartesian form: x + y + z +17 = 0 The point is $(\widehat i + 2\widehat j + 5\widehat k)$ => (1, 2,...
from the plane 
.Answer: Given plane, $\overrightarrow r .(3\widehat i - 4\widehat j + 12\widehat k) = 9$ The cartesian form: 3x – 4y + 12z – 9 =0 The point is $(2\widehat i - \widehat j - 4\widehat k)$ => (2,...
Answer: Given, A = (1, 4, 6)
Answer: Given, d = 6
Answer: Given, = (x × 2) + (y × (-3)) + (z × 4) = 2x - 3y + 4z The Cartesian equation...
Answer: Given, $\begin{array}{l} d = 5\\ \widehat n = \widehat k \end{array}$ The equation of plane at 5 units distance from the origin $\begin{array}{l} \widehat n \end{array}$ and as a unit...
Answer: Equation of the plane: 4x – 3y + 2z = 12 $\begin{array}{l} \frac{4}{{12}}x - \frac{3}{{12}}y + \frac{2}{{12}}z = 1\\ \frac{x}{3} + \frac{y}{{ - 4}} + \frac{z}{6} = 1 \end{array}$ It is the...
Answer: Given, Coordinate axes are 2, - 4, 5 The equation of the variable plane: The required equation of the plane is 10x – 5y + 4z = 20.
Answer: Given, A (0, -1, 0) B (2, 1, -1) C (1, 1, 1) D (3, 3, 0) 4x – 3 (y + 1) + 2z = 0 4x – 3y + 2z – 3 = 0 Take x = 0, y = 3 and z =...
Answer: Given, A (-2, 6, -6) B (-3, 10, -9) C (-5, 0, -6) ...
Answer: Let us take, The equation of the plane passing through A (3, 2, -5) a (x – 3) + b (y – 2) + c (z + 5) = 0 It passes through the points B (-1, \4, -3) and C (-3, 8, -5) a (1 – 3) + b (4 – 2)...
Answer: (i) Given, A (2, 2, -1) B (3, 4, 2) C (7, 0, 6) ...
Solution: $=\left|\begin{array}{lll} b(b-a) & b-c & c(b-a) \\ a(b-a) & a-b & b(b-a) \\ c(b-a) & c-a & a(b-a) \end{array}\right|$ Taking (b-a) common from $\mathrm{C}_{1},...
Solution: Expanding with R1 $\begin{array}{l} =b^{2} c^{2}\left(a^{2} c+a b c-a b c-a^{2} b\right)-b c\left(a^{3} c^{2}+a^{2} b c^{2}-a^{2} b^{2} c-a^{3} b^{2}\right)+(b+c)\left(a^{3} b c^{2}-a^{3}...
Solution: $\left|\begin{array}{ccc}a^{2} & b^{2} & c^{2} \\ (a+1)^{2} & (b+1)^{2} & (c+1)^{2} \\ (a-1)^{2} & (b-1)^{2} & (c-1)^{2}\end{array}\right|$...
Solution: $\left|\begin{array}{ccc} a & b & a x+b y \\ b & c & b x+c y \\ a x+b y & b x+c y & 0 \end{array}\right|$ $\begin{array}{l} \left.=\left(\frac{1}{x...
Solution: $\begin{array}{l} \left|\begin{array}{ccc} \mathrm{b}+\mathrm{c} & \mathrm{a} & \mathrm{a} \\ \mathrm{b} & \mathrm{c}+\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{c}...
Solution: $\begin{array}{l} \left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| \\ =\left|\begin{array}{ccc} a^{2}-1 & a-1...
Solution: $\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|$ $=\left|\begin{array}{ccc}5 \mathrm{x}+4 & 5 \mathrm{x}+4...
Solution: $\begin{array}{l} \left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right| \\...
is a 
matrix such that 
and 
then write the value of 
.Solution: Theorem: If Let $A$ be $k \times k$ matrix then $|p A|=p^{k}|A|$. Given: $\mathrm{k}=3$ and $\mathrm{p}=3$. $\begin{array}{l} |3 \mathrm{~A}|=3^{3} \times|\mathrm{A}| \\ =27|\mathrm{~A}|...
Answer:
Answer: = cosx cos2x cos4x cos8x Multiply and divide by 2sinx, we get We know that, sin 2x = 2 sinx cosx Replacing x by 2x, we get sin 2(2x) = 2 sin(2x) cos(2x) or sin 4x = 2 sin 2x cos 2x...
Answer: Taking sinx common from the numerator and cosx from the denominator
Answer: = RHS ∴ LHS = RHS Hence Proved
Answer: Multiply and divide by 2, we get = RHS ∴ LHS = RHS Hence Proved
Answer: Taking LHS, = cot x – 2cot 2x …(i) We know that, cot x = cos x/ sin x Replacing x by 2x, we get cot 2x = cos 2x/ sin 2x So, eq. (i) becomes = sin x/ cos x = tan x = RHS ∴ LHS = RHS Hence...
Answer: Taking LHS = 2 – 1 = 1 = RHS
Answer: To Prove: cosec 2x + cot 2x = cot x Taking LHS, = cosec 2x + cot 2x …(i) We know that, cosec x = 1 / sin x and cot x = cos x/ sin x Replacing x by 2x, we get = cos x/ sinx = cot...
Answer: Taking LHS, sin 2x(tan x + cot x) We know that: We know that, sin 2x = 2 sinx cosx = 2 = RHS ∴ LHS = RHS Hence Proved
Answer: = sin x / cos x = tan x = RHS ∴ LHS = RHS Hence Proved
Sol: Given: Base = 25 cm Height = 16.8 cm \Area of the parallelogram = Base ´ Height = 25cm ´16.8 cm = 420 cm2
and 
bulbs respectively. Out of these bulbs 
and 
of the bulbs produced respectively by 
and 
are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by machine .Let $A$ : Manufactured from machine $A$, B : Manufactured from machine B C: Manufactured from machine C D : Defective bulb We want to find $P(A \mid D)$, i.e. probability of selected defective bulb...
, and that of motorcycles is 
. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.Let $M$ : Motorcycle S: Scooter A : Accident vechicle We want to find $P(M \mid A)$, i.e. probability of accident vehicle was a motorcycle $\begin{array}{l} P(M \mid A)=\frac{P(M) \cdot P(A \mid...
and 
. Plant manufactures 
of the cars, and plant , manufactures 
. At pant 
of the cars are rated of standard quality, and at plant 
are rated of standard quality. A car is picked up at random and is found to be of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant .Let $X$ : Car produced from plant $X$ $Y$ : Car produced from plant $Y$ S: Car rated as standard quality We want to find $P(X \mid S)$, i.e. selected standard quality car is from plant $X$...
and 
, containing marbles. A contains 1 red, 6 white and 3 black marbles; 
contains 6 red, 2 white and 2 black marbles; C contains 8 red, 1 white and 1 black marbles; and D contains 6 white and 4 black marbles. One of the boxes is selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from the box ?Let $A:$ Ball drawn from bag $A$ B: Ball is drawn from bag B $C:$ Ball is drawn from bag $C$ $D:$ Ball is drawn from bag $D$ BB: Black ball WB : White ball RB : Red ball Assuming all boxes have an...
Let $A$ : the set of first 3 bags $B$ : a set of next 2 bags WB : White ball BB : Black ball Now we can change the problem to two bags, i.e. bag A containing 15 white and 9 black balls( 5 white and...
and 
respectively. From the chosen urn, two balls are drawn at random without replacement. Both the balls happen to be white. Find the probability that the balls are drawn are from urn .Let $A:$ Ball is drawn from bag $A$ B : Ball is drawn from bag B $C:$ Ball is drawn from bag $C$ BB: Black ball WB: White ball RB: Red ball Probability of picking 2 white balls fro urn $A=\frac{7...
let $A:$ Ball drawn from bag $A$ B : Ball is drawn from bag B $C:$ Ball is drawn from bag $C$ BB: Black ball WB: White ball RB: Red ball Assuming, selecting bags is of equal probability i.e....
Solution: Option(C) is correct. To Find: The value of $\tan ^{-1} 2+\tan ^{-1} 3$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $\begin{array}{l}...
let $\mathrm{A}:$ Ball drawn from bag $\mathrm{A}$ $B:$ Ball is drawn from bag $B$ $C:$ Ball is drawn from bag $C$ BB : Black ball WB : White ball Assuming, selecting bags is of equal probability...
.let $A:$ Ball drawn from bag $A$ B : Ball is drawn from bag B $C:$ Ball is drawn from bag $C$ R: Red ball W : White ball Assuming, selecting bags is of equal probability i.e. $\frac{1}{3}$ We want...
Let $\mathrm{W}$ : White ball B : Black ball $\begin{array}{l} X: 1^{\text {st }} \text { bag } \\ Y: 2^{\text {nd }} \text { bag } \end{array}$ Assuming, selecting bags is of equal probability i.e....
. red balls. Another bag contains 4 white and 3 red balls. One of the bags is selected at random, and a ball is drawn from it, which is found to be white. Find the probability that the ball is drawn is from bag A.Let R : Red ball W : White ball A: Bag A B: Bag B Assuming, selecting bags is of equal probability i.e. $\frac{1}{2}$ We want to find $P(A \mid W)$, i.e. the selected white ball is from bag $A$:...
and 
, respectively. Further, if the first group wins, the probability of introducing a new product is 
, and when the second groups win, the corresponding probability is 
. Find the probability that the new product introduced was by the second group.Let $F$ : First group S : Second group $N$ : Introducing a new product We want to find $P(S \mid N)$, i.e. new product introduced by the second group $\begin{array}{l} \mathrm{P}(\mathrm{S} \mid...
Solution: Option(C) is correct. To Find: The value of $\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}$ Let, $x=\tan ^{-1} 1+\tan ^{-1} \frac{1}{3}$ Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan...
Solution: Option(A) is correct. To Find: The value of $\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ Let, $x=\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ $\Rightarrow...
Solution: Option(B) is correct. To Find: The value of $\tan ^{-1} 1+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$ Now, let $x=\tan ^{-1} 1+\cos...
Solution: Option(B) is correct. To Find: The value of $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$ Let, $x=\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$ $\begin{array}{l} \Rightarrow...
is 
Solution: Option(B) is correct. To Find: The value of $\sin \left(\cos ^{-1} \frac{3}{5}\right)$ Now, let $x=\cos ^{-1} \frac{3}{5}$ $\Rightarrow \cos x=\frac{3}{5}$ Now , $\sin x=\sqrt{1-\cos ^{2}...
is 
Solution: Option(A) is correct. To Find: The value of $\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ Now, let $x=\sec ^{-1}\left(\sec \left(\frac{8 \pi}{5}\right)\right)$ $\Rightarrow...
is 
Solution: To Find: The value of $\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$ Now, let $x=\tan ^{-1}\left(\tan \left(\frac{7 \pi}{6}\right)\right)$ $\Rightarrow \tan x=\tan...
is Solution: Option(C) us correct. To Find: The value of $\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ Now, let $x=\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ $\Rightarrow...
is 
Solution: Option(A) is correct. To Find: The Principle value of $\operatorname{cosec}^{-1}(-\sqrt{2})$ Let the principle value be given by $\mathrm{x}$ Now, let...
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![\[\frac{3}{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b506a58aabfe65c0a3f19affd6f9c8e7_l3.png "Rendered by QuickLaTeX.com")
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Given Eccentricity = \[\frac{3}{4}\] We know that Eccentricity = c/a Therefore,c=\[\frac{3}{4}\]a
![\[\left( \mathbf{4},\text{ }\mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6e14255e5f2d2cbc940379e069bb5844_l3.png "Rendered by QuickLaTeX.com")
![\[(-1,4)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-aeaab2e61affb979c28d3883772be7c5_l3.png "Rendered by QuickLaTeX.com")
Given: Center is at the origin and Major axis is along x – axis So, Equation of ellipse is of the form \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]…(i) Given that ellipse passing...
![\[\left( \mathbf{0},\text{ }\pm \mathbf{4} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bb2be58eb8036aac914c2c677f788949_l3.png "Rendered by QuickLaTeX.com")
![\[e=\frac{4}{5}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c0fb8f8c87448e0d2f359f2d167e4c1a_l3.png "Rendered by QuickLaTeX.com")
Given: Coordinates of foci = \[\left( \mathbf{0},\text{ }\pm \mathbf{4} \right)\] …(i) We know that, Coordinates of foci = \[\left( 0,\text{ }\pm c \right)\] …(ii) The coordinates of the foci are...
![\[\left( \pm \mathbf{1},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-873e68902b5445bb221bebb969489511_l3.png "Rendered by QuickLaTeX.com")
Let the equation of the required ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Given: Coordinates of foci = \[\left( \pm 1,\text{ }0 \right)\] …(i) We know that,...
![\[\left( \pm \mathbf{2},\text{ }\mathbf{0} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7120150033b6ca5ebc01c838e0e9b285_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{1}{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-eca8ad2eb4fe01867f7756c04c9a2882_l3.png "Rendered by QuickLaTeX.com")
Let the equation of the required ellipse be Given: Coordinates of foci = \[\left( \pm 2,\text{ }0 \right)\]…(iii) We know that, Coordinates of foci = \[\left( \pm c,\text{ }0 \right)\]…(iv) ∴ From...
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![\[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2d6b922fc27485fd054fb582ccf195be_l3.png "Rendered by QuickLaTeX.com")
Given: Ends of Major Axis = \[\left( \pm \mathbf{4},\text{ }\mathbf{0} \right)\] and Ends of Minor Axis = \[\left( \mathbf{0},\text{ }\pm \mathbf{3} \right)\] Here, we can see that the major axis is...
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![\[(0,\pm \sqrt{7})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1992b2f60ca3bf6ba810d7743190c38a_l3.png "Rendered by QuickLaTeX.com")
Given: Vertices = \[(0,\pm 4)\] …(i) The vertices are of the form = (0, ±a) …(ii) Hence, the major axis is along y – axis ∴ From eq. (i) and (ii), we get \[\begin{array}{*{35}{l}} a\text{ }=\text{...
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![\[(\pm \mathbf{4},~\mathbf{0})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-10ca5316ffee057110a312dedd039adb_l3.png "Rendered by QuickLaTeX.com")
Given: Vertices = \[\left( \pm \mathbf{6},\text{ }\mathbf{0} \right)\] …(i) The vertices are of the form = \[\left( \pm a,\text{ }0 \right)\] …(ii) Hence, the major axis is along x – axis ∴ From eq....
![\[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3badc0893d1a72abf2537d836a63e857_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
Given: \[\mathbf{25}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{4}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{100}\] Divide by \[100\] to both the sides, we get...
![\[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c1bda8518d8de0566fb7bb8dae83e7d6_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}\] Divide by \[16\] to both the sides, we get...
Answer: = cot x = RHS ∴ LHS = RHS Hence Proved
![\[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-126ca6c456d3168070fa6d14075b510f_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}\] Divide by \[36\] to both the sides, we get \[\frac{9}{36}{{x}^{2}}+\frac{1}{36}{{y}^{2}}=1\]...
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}\] Divide by \[36\] to both the sides, we get \[\frac{9}{36}{{x}^{2}}+\frac{1}{36}{{y}^{2}}=1\]...
Given: \[\mathbf{9}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{36}\] Divide by \[36\] to both the sides, we get \[\frac{9}{36}{{x}^{2}}+\frac{1}{36}{{y}^{2}}=1\]...
Answer: = tan x = RHS ∴ LHS = RHS Hence Proved
![\[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-05cca57fba353b302ed103f21033d9a8_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}\]…(i) Divide by \[18\] to both the sides, we get...
Answer: Taking LHS = cos x + sin x = RHS ∴ LHS = RHS Hence Proved
Given: \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}\]…(i) Divide by \[18\] to both the sides, we get...
Given: \[\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{2}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{18}\]…(i) Divide by \[18\] to both the sides, we get...
Answer; We know that, cos 3x = 4cos3 x – 3 cosx Putting the values, we get
Answer: To find: sin 3x We know that, sin 3x = 3 sinx – sin3 x Putting the values, we get
![\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1f0a66ca234bf1c2806a47a69c3bf010_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing eq. (i)...
Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing...
Given: \[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{16}=1\]….(i) Since, \[9<16\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii) Comparing eq. (i)...
, find the values of tan 2xAnswer: We know that:
Given: \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]…(i) Since, \[4\text{ }<\text{ }25\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii)...
, find the values of cos 2xAnswer:
![\[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-2cd71d027e4ee356834184fd92e4f618_l3.png "Rendered by QuickLaTeX.com")
Given: \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{25}=1\]…(i) Since, \[4\text{ }<\text{ }25\] So, above equation is of the form, \[\frac{{{x}^{2}}}{{{b}^{2}}}+\frac{{{y}^{2}}}{{{a}^{2}}}=1\]…(ii)...
, find the values of sin 2xAnswer: We know that
![\[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c30dd6bfab0df4e91e525631f2ba2f4a_l3.png "Rendered by QuickLaTeX.com")
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since,...
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since,...
Given: \[\mathbf{4}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }\mathbf{9}{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{1}\] \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]….(i) Since, ...
, find the values of tan 2xAnswer: Replacing x by 2x, we get tan 2x = sin 2x / cos 2x Putting the values of sin 2x and cos 2x, we get
, find the values of cos 2xAnswer: We know that, cos 2x = 2cos2 x – 1 Putting the value, we get
, find the values of sin 2xAnswer: We know that, sin2x = 2 sinx cosx …(i) Here, we don’t have the value of sin x. So, firstly we have to find the value of sinx We know that, cos2 x + sin2 x = 1 Putting the values, we get...
, find the values of tan 2xAnswer: We know that: Replacing x by 2x, we get tan 2x = sin 2x / cos 2x Putting the values of sin 2x and cos 2x, we get
, find the values of cos 2xAnswer: We know that, cos 2x = 1 – 2sin2 x Putting the value, we get
, find the values of sin 2xAnswer: Given: $\sin x=\frac{\sqrt{5}}{3}$ To find: sin2x We know that, sin2x = 2 sinx cosx …(i) Here, we don’t have the value of cos x. So, firstly we have to find the value of cosx We know that,...