cos left(sin ^{-1} frac{1}{5}+cos ^{-1} xrigh | vedclass

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

Question

(B) Given the equation: $\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} xight)=0$. We know that $\cos 	heta = 0$ implies $	heta = \frac{\pi}{2}$ (co

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(B) Given the equation: $\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} xight)=0$. We know that $\cos 	heta = 0$ implies $	heta = \frac{\pi}{2}$ (considering the principal value branch). Therefore,$\sin ^{-1} \frac{1}{5}+\cos ^{-1} x = \frac{\pi}{2}$. Rearranging the terms,we get $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} \frac{1}{5}$. Using the identity $\sin ^{-1} y + \cos ^{-1} y = \frac{\pi}{2}$,we have $\cos ^{-1} y = \frac{\pi}{2} - \sin ^{-1} y$. Thus,$\cos ^{-1} x = \cos ^{-1} \frac{1}{5}$. Comparing both sides,we get $x = \frac{1}{5}$.

$\cos \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=0$,then $x=$ . . . . . . .

Similar Questions

Evaluate: $\tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3}) = $ . . . . . . .

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2},$ then the value of $x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}}$ is equal to

The value of $\cot \left(\sum_{n=1}^{23} \cot ^{-1}\left(1+\sum_{k=1}^n 2 k\right)\right)$ is

$\tan \left[ \cos^{-1} \frac{4}{5} + \tan^{-1} \frac{2}{3} \right] =$

Find $\frac{dy}{dx}$ if $y = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$,where $0 < x < \frac{1}{\sqrt{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module