If 4 cos^{-1} x + sin^{-1} x = frac{pi}{2},th | vedclass

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

Question

(A) We are given the equation: $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$.We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,which i

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We are given the equation: $4 \cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$.We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,which implies $\sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x$.Substituting this into the given equation:$4 \cos^{-1} x + (\frac{\pi}{2} - \cos^{-1} x) = \frac{\pi}{2}$.Simplifying the expression:$3 \cos^{-1} x + \frac{\pi}{2} = \frac{\pi}{2}$.Subtracting $\frac{\pi}{2}$ from both sides:$3 \cos^{-1} x = 0$.Dividing by $3$:$\cos^{-1} x = 0$.Taking the cosine of both sides:$x = \cos(0) = 1$.

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

Similar Questions

If $\tan ^{-1} x+\tan ^{-1} y=\frac{4 \pi}{5}$,then $\cot ^{-1} x+\cot ^{-1} y$ is equal to

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

If $\sin^{-1} x = \theta + \beta$ and $\sin^{-1} y = \theta - \beta$,then $1 + xy = $

$\cos \left[ {{\cos }^{ - 1}}\left( {\frac{{ - 1}}{7}} \right) + {{\sin }^{ - 1}}\left( {\frac{{ - 1}}{7}} \right) \right] = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module