Let f(theta) = sin ( tan ^{-1} ( frac{sin the | vedclass

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(B) Given $f(	heta) = \sin ( 	an ^{-1} ( \frac{\sin 	heta}{\sqrt{\cos 2 	heta}} ) )$. Using the identity $	an ^{-1} x = \sin ^{-1} ( \frac{x}{\sq

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(B) Given $f(	heta) = \sin ( 	an ^{-1} ( \frac{\sin 	heta}{\sqrt{\cos 2 	heta}} ) )$. Using the identity $	an ^{-1} x = \sin ^{-1} ( \frac{x}{\sqrt{1+x^2}} )$,we have: $f(	heta) = \sin ( \sin ^{-1} ( \frac{\frac{\sin 	heta}{\sqrt{\cos 2 	heta}}}{\sqrt{1 + \frac{\sin ^2 	heta}{\cos 2 	heta}}} ) )$ $f(	heta) = \sin ( \sin ^{-1} ( \frac{\sin 	heta}{\sqrt{\cos 2 	heta + \sin ^2 	heta}} ) )$ Since $\cos 2 	heta = \cos ^2 	heta - \sin ^2 	heta$,we get: $f(	heta) = \sin ( \sin ^{-1} ( \frac{\sin 	heta}{\sqrt{\cos ^2 	heta - \sin ^2 	heta + \sin ^2 	heta}} ) )$ $f(	heta) = \sin ( \sin ^{-1} ( \frac{\sin 	heta}{\sqrt{\cos ^2 	heta}} ) )$ For $-\frac{\pi}{4}  0$,so $\sqrt{\cos ^2 	heta} = \cos 	heta$. $f(	heta) = \sin ( \sin ^{-1} ( \frac{\sin 	heta}{\cos 	heta} ) ) = \sin ( \sin ^{-1} ( 	an 	heta ) ) = 	an 	heta$. Now,we need to find $\frac{d}{d(	an 	heta)}(f(	heta))$. Let $u = 	an 	heta$. Then $f(	heta) = u$. Therefore,$\frac{d}{du}(u) = 1$.

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

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