For -1

Question

(A) Given the function $f(x) = \cos^2 \left( 	an^{-1} \sqrt{\frac{1-x}{1+x}} ight)$ for $-1 < x < 1$. Let $x = \cos(2	heta)$,where $0 < 2	heta <

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given the function $f(x) = \cos^2 \left( 	an^{-1} \sqrt{\frac{1-x}{1+x}} ight)$ for $-1 Let $x = \cos(2	heta)$,where $0 Then,$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos(2	heta)}{1+\cos(2	heta)}} = \sqrt{\frac{2\sin^2	heta}{2\cos^2	heta}} = 	an	heta$. Substituting this into the function,we get $f(x) = \cos^2(	an^{-1}(	an	heta)) = \cos^2	heta$. Using the identity $\cos^2	heta = \frac{1+\cos(2	heta)}{2}$,we have $f(x) = \frac{1+x}{2}$. Differentiating with respect to $x$,we get $f'(x) = \frac{d}{dx} \left( \frac{1}{2} + \frac{x}{2} ight) = \frac{1}{2}$.

For $-1 < x < 1$,if $f(x) = \cos^2 \left( \tan^{-1} \sqrt{\frac{1-x}{1+x}} \right)$,then $f'(x) =$

Similar Questions

If $y = \frac{K^{\cos^{-1} x}}{1 + K^{\cos^{-1} x}}$ and $t = K^{\cos^{-1} x}$,then find $\frac{dy}{dt}$.

If $y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$,then $\frac{dy}{dx} = \ldots$

The derivative of $\sec^{-1}\left( \frac{1}{2x^2 - 1} \right)$ with respect to $\sqrt{1 - x^2}$ at $x = \frac{1}{2}$ is:

Let $f(\theta) = \sin \left(\tan^{-1} \left(\frac{\sin \theta}{\sqrt{\cos 2\theta}} \right) \right)$,where $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$. Then the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module