^ -1 (1 2 ) 1. 4 2. 3 3. 6 4. 2 - Vidyadip

Question

$\sin^{-1}\Big(\frac{1}{\sqrt2}\Big)$

$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
$\frac{\pi}{2}$

✓

Answer

$\frac{\pi}{4}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from INVERSE TRIGNOMETRIC FUNCTIONS→INVERSE TRIGNOMETRIC FUNCTIONS — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

$\cos ^{-1}\left(\frac{-1}{2}\right)+2 \sin ^{-1}\left(\frac{-1}{2}\right)$ is equal to

Out of the given matrices, choose that matrix which is a scalar matrix:

$\begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\begin{bmatrix}0&0&0\\0&0&0\end{bmatrix}$
$\begin{bmatrix}0&0\\0&0\\0&0\end{bmatrix}$
$\begin{bmatrix}0\\0\\0\end{bmatrix}$

What is integrating factor of $\frac{\text{dy}}{\text{dx}}+\text{y}\sec\text{x}=\tan\text{x}?$

$\sec\text{x}+\tan\text{x}$
$\log(\sec\text{x}+\tan\text{x})$
$\text{e}^{\sec\text{x}}$
$\sec{\text{x}}$

The direction cosines l, m and n of two lines are connected by the relations l + m + n = 0, l m = 0, then the angles between them is:

$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
0

If $\tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\text{A},$ then A is equal to:

$\text{x}-\text{y}$
$\text{x}+\text{y}$
$\frac{\text{x}-\text{y}}{1+\text{xy}}$
$\frac{\text{x}+\text{y}}{1-\text{xy}}$

Find the minor of the element 1 in the determinant $\triangle=\begin{bmatrix}1&5\\3&8\end{bmatrix}$ is:

5
1
8
3

If the fucnction $\text{f(x)}=\begin{cases}(\cos\text{x})^{\frac{1}{\text{x}}},&\text{x}\neq0\\\text{k},&\text{x}=0\end{cases}$ is continuouse at x = 0, then the value of k is:

0
1
-1
e

If $\sin\text{y}=\text{x}\cos(\text{a}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}$ is equal to:

$\frac{\cos^2(\text{a}+\text{y})}{\cos\text{a}}$
$\frac{\cos\text{a}}{\cos^2(\text{a}+\text{y})}$
$\frac{\sin^2\text{y}}{\cos\text{a}}$
$\text{None of these.}$

The equation of the line passing through the points $\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$ and $\text{b}\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$ is:

$\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)+\lambda\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$\vec{\text{r}}=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big)-\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$\vec{\text{r}}=\text{a}_1(1-\text{t})\hat{\text{i}}+\text{a}_2(1-\text{t})\hat{\text{j}}+\text{a}_3(1-\text{t})\hat{\text{k}}+\text{t}\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$\text{None of these}$

If $\alpha=\tan^{-1}\Big(\frac{\sqrt3\text{x}}{2\text{y}-\text{x}}\Big),\beta=\tan^{-1}\Big(\frac{2\text{x}-\text{y}}{\sqrt3\text{y}}\Big),$ then $\alpha-\beta=$

$\frac{\pi}{6}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$
$-\frac{\pi}{3}$