If sin ^{-1} frac{x}{5}+sin ^{-1} frac{4}{5}= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation: $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$ We know that $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$ for $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Given the equation: $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$ We know that $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$ for $y \in [-1, 1]$. Thus,$\sin ^{-1} \frac{x}{5} = \frac{\pi}{2} - \sin ^{-1} \frac{4}{5}$. Using the identity,we get $\sin ^{-1} \frac{x}{5} = \cos ^{-1} \frac{4}{5}$. Let $\cos ^{-1} \frac{4}{5} = 	heta$,then $\cos 	heta = \frac{4}{5}$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$. Therefore,$\sin ^{-1} \frac{x}{5} = \sin ^{-1} \frac{3}{5}$. Comparing both sides,we get $\frac{x}{5} = \frac{3}{5}$,which implies $x = 3$.

If $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$,then $x=$ . . . . . . .

Similar Questions

$\tan ^{-1} 2+\tan ^{-1} 3=$

The solution of the equation $\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$ is

If $y = \sin^{-1} \left( \frac{\sqrt{1+x} + \sqrt{1-x}}{2} \right)$,then $\frac{dy}{dx} = $

If $y = \sin^{-1}\left(\frac{x^2 - 1}{x^2 + 1}\right) + \sec^{-1}\left(\frac{x^2 + 1}{x^2 - 1}\right)$,$|x| > 1$,then $\frac{dy}{dx}$ is equal to:

Evaluate: $\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module