Question
If $\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\text{x}=\frac{\pi}{2},$ find x.

Answer

We know that $\sin^{-1}\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$

We have

$\sin^{-1}\Big(\frac{1}{3}\Big)+\cos^{-1}\text{x}=\frac{\pi}{2}$

$\Rightarrow\sin^{-1}\Big(\frac{1}{3}\Big)=\frac{\pi}{2}-\cos^{-1}\text{x}$

$\Rightarrow\sin^{-1}\Big(\frac{1}{3}\Big)=\sin^{-1}\text{x}$

$\Rightarrow\text{x}=\frac{1}{3}$

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

Similar questions

Evaluate the following integrals:
$\int\Big(\frac{\text{m}}{\text{x}}+\frac{\text{x}}{\text{m}}+\text{m}^\text{x}+\text{x}^\text{m}+\text{mx}\Big)\text{dx}$
If a particle moves in a straight line such that the distance travelled in time t is given by s = t3 - 6t2 + 9t + 8. Find the initial velocity of the particle.
Solve:
$5\tan^{-1}\text{x}+3\cot^{-1}\text{x}={2\pi}$
Determine whether the following operations define a binary operation on the given set or not:
6' on S = {1, 2, 3, 4, 5} defined by, a ×6 b = Remainder when ab is divided by 6.
Find the principal values:
 $\cos^{-1}\bigg(-\frac{1}{\sqrt{2}}\bigg)$
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Of the same colour.
If $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0,$ what can you conclude about the vector $\vec{\text{b}}$?
If x and y are connected parametrically by the equation x = a ($\theta$ – sin $\theta$), y = a (1 + cos $\theta$) without eliminating the parameter. Find $\frac{d y}{d x}$.
If $\text{f}'\text{(x)}=\text{x}-\frac{1}{\text{x}^2}$ and $\text{f}'(1)=\frac{1}{2}$, find f'(x).
Find the adjoint of the following matrices:

$\text{C}=\begin{bmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$

Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.