If sin ^{-1}left(frac{12}{x}right)+sin ^{-1}l | vedclass

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

Question

(C) Given equation is $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$.We know that $\sin ^{-1}(u) + \cos ^{-1}(

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Given equation is $\sin ^{-1}\left(\frac{12}{x}ight)+\sin ^{-1}\left(\frac{5}{x}ight)=\frac{\pi}{2}$.We know that $\sin ^{-1}(u) + \cos ^{-1}(u) = \frac{\pi}{2}$, so $\sin ^{-1}(u) = \frac{\pi}{2} - \cos ^{-1}(u)$.Rearranging the given equation: $\sin ^{-1}\left(\frac{12}{x}ight) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{5}{x}ight)$.Using the identity $\frac{\pi}{2} - \sin ^{-1}(u) = \cos ^{-1}(u)$, we get $\sin ^{-1}\left(\frac{12}{x}ight) = \cos ^{-1}\left(\frac{5}{x}ight)$.Let $\sin ^{-1}\left(\frac{12}{x}ight) = 	heta$, then $\sin 	heta = \frac{12}{x}$.Since $\sin^2 	heta + \cos^2 	heta = 1$, we have $\cos 	heta = \sqrt{1 - \left(\frac{12}{x}ight)^2} = \frac{\sqrt{x^2 - 144}}{x}$.Thus, $	heta = \cos ^{-1}\left(\frac{\sqrt{x^2 - 144}}{x}ight)$.Equating the arguments: $\frac{\sqrt{x^2 - 144}}{x} = \frac{5}{x}$.Squaring both sides: $x^2 - 144 = 25$.$x^2 = 169$, which gives $x = 13$ (since $x$ must be positive for the domain of $\sin^{-1}$ here).

If $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$, then $x=$

Similar Questions

If $y = \sin^{-1} \left( \frac{5x + 12\sqrt{1-x^2}}{13} \right)$,then $\frac{dy}{dx} = $

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

$\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = $

If $\sin^{-1} x + \sin^{-1} y = \pi/2$, then $x^2$ is equal to

If $\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module