If sin ^{-1}left(frac{3}{x}right)+sin ^{-1}le | vedclass

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

Question

(B) Given that, $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$.We know that $\sin ^{-1}(A) + \cos ^{-1}(A) = \f

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Given that, $\sin ^{-1}\left(\frac{3}{x}ight)+\sin ^{-1}\left(\frac{4}{x}ight)=\frac{\pi}{2}$.We know that $\sin ^{-1}(A) + \cos ^{-1}(A) = \frac{\pi}{2}$, so $\sin ^{-1}\left(\frac{4}{x}ight) = \frac{\pi}{2} - \cos ^{-1}\left(\frac{4}{x}ight)$.Rearranging the given equation: $\sin ^{-1}\left(\frac{3}{x}ight) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{x}ight)$.Using the identity $\cos ^{-1}(A) = \frac{\pi}{2} - \sin ^{-1}(A)$, we get $\sin ^{-1}\left(\frac{3}{x}ight) = \cos ^{-1}\left(\frac{4}{x}ight)$.Let $\sin ^{-1}\left(\frac{3}{x}ight) = 	heta$, then $\sin 	heta = \frac{3}{x}$, which implies $\cos 	heta = \sqrt{1 - \left(\frac{3}{x}ight)^2} = \frac{\sqrt{x^2-9}}{x}$.Thus, $	heta = \cos ^{-1}\left(\frac{\sqrt{x^2-9}}{x}ight)$.Equating the two expressions: $\frac{\sqrt{x^2-9}}{x} = \frac{4}{x}$.Squaring both sides: $x^2 - 9 = 16$, which gives $x^2 = 25$.Since the domain of $\sin ^{-1}$ requires $|\frac{3}{x}| \le 1$ and $|\frac{4}{x}| \le 1$, we have $|x| \ge 4$. Thus, $x = 5$.

If $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$, then $x$ is equal to

Similar Questions

$2 \tan^{-1}(\cos x) = \tan^{-1}(\csc^2 x)$,then $x = $

$4 \tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$ is equal to

$2 \pi - \left(\sin ^{-1} \frac{4}{5} + \sin ^{-1} \frac{5}{13} + \sin ^{-1} \frac{16}{65}\right)$ is equal to

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:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module