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 <

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$\frac{1}{2}$

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 $f'(x) =$

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