Class 10

### 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.

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...

### If then A. B. C. D.

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.

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: A. B. C. D.

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: 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: 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...

### Find the (v) length of the latus rectum of each of the following ellipses.

Given: $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$ Divide by $16$ to both the sides, we get...

### Find the (i) lengths of major axes, (ii) coordinates of the vertices

Given: $\mathbf{16}{{\mathbf{x}}^{\mathbf{2}}}+\text{ }{{\mathbf{y}}^{\mathbf{2}}}=\text{ }\mathbf{16}$ Divide by $16$ to both the sides, we get...

### Find the (iii) coordinates of the foci, (iv) eccentricity

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)...

### Mark the tick against the correct answer in the following: The principal value of is A. B. C. D.

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 $x=\sin ^{-1}\left(\frac{-1}{2}\right)$...

### Write down the interval for the principal-value branch of each of the following functions and draw its graph:

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:

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:

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:

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:

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:

Solution: Principal value branch of $\sin ^{-1} x$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

### 6. Construct a ABC in which AB = 6 cm, A  30and AB  60 . Construct

Sol: Steps of Construction with base AB’ = 8 cm. Step 1: Draw a line segment AB = 6cm. Step 2: At A, draw ÐXAB = 30°. Step 3: At B, draw ÐYBA = 60°. Suppose AX and BY intersect at C. Thus, DABC is...

### Solve for :

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 x:

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: $\tan ^{-1}(2+x)+\tan ^{-1}(2-x)=\tan ^{-1} \frac{2}{3}$...

### Prove that:

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:

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:

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) $1+\cos A=2 \cos ^{2} \frac{A}{2}$...

### Find the principal value of each of the following :

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 :

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 :

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 :

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 :

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 :

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 :

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 :

Solution: $\operatorname{cosec}^{-1}(2)$ Putting the value directly $=\frac{\pi}{6}$

Solution: $\begin{array}{l} \text { i) } \vec{a}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k} \\ \vec{b}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \mathrm{I} \vec{a}... read more ### Write the projection of vector along the vector . Solution:$\begin{array}{l} \vec{a}=\hat{\imath}+\hat{\jmath}+\hat{k} \\ \vec{b}=\hat{\jmath} \end{array}\mathrm{I} \vec{a} \mathrm{I}=\sqrt{3}=\sqrt{1^{2}+1^{2}+1^{2}}\begin{array}{l}...
Solution: If $\vec{a}=\mathrm{a} 1 \hat{\imath}+\mathrm{a} 2 \hat{\jmath}+\mathrm{a}_{3} \hat{k}$ and $\vec{b}=\mathrm{b}_{1} \hat{\imath}+\mathrm{b}_{2} \hat{\jmath}+\mathrm{b}_{3} \hat{k}$, then...