If f(x) = frac{sin^{2} x}{1+cot x} + frac{cos | vedclass

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(C) Given,$f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+	an x}$.Simplify the expression:$f(x) = \frac{\sin^{2} x}{1+\frac{\cos x}{\sin x}

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(C) Given,$f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+	an x}$.Simplify the expression:$f(x) = \frac{\sin^{2} x}{1+\frac{\cos x}{\sin x}} + \frac{\cos^{2} x}{1+\frac{\sin x}{\cos x}}$$f(x) = \frac{\sin^{3} x}{\sin x + \cos x} + \frac{\cos^{3} x}{\cos x + \sin x}$$f(x) = \frac{\sin^{3} x + \cos^{3} x}{\sin x + \cos x}$Using the identity $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2})$:$f(x) = \frac{(\sin x + \cos x)(\sin^{2} x - \sin x \cos x + \cos^{2} x)}{\sin x + \cos x}$$f(x) = \sin^{2} x - \sin x \cos x + \cos^{2} x$$f(x) = 1 - \sin x \cos x = 1 - \frac{1}{2} \sin(2x)$.Now,differentiate with respect to $x$:$f^{\prime}(x) = -\frac{1}{2} \cdot \cos(2x) \cdot 2 = -\cos(2x)$.Substitute $x = \frac{\pi}{4}$:$f^{\prime}\left(\frac{\pi}{4}ight) = -\cos\left(2 \cdot \frac{\pi}{4}ight) = -\cos\left(\frac{\pi}{2}ight) = 0$.

If $f(x) = \frac{\sin^{2} x}{1+\cot x} + \frac{\cos^{2} x}{1+\tan x}$,then $f^{\prime}\left(\frac{\pi}{4}\right)$ is

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