If f(x) = frac{sin^2 x}{1+cot x} + frac{cos^2 | 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 terms:$f(x) = \frac{\sin^2 x}{1+\frac{\cos x}{\sin x}} + \frac{\

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$-\frac{1}{2}$

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(C) Given $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+	an x}$.Simplify the terms:$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}$$f(x) = \sin^2 x + \cos^2 x - \sin x \cos x$$f(x) = 1 - \frac{1}{2}(2 \sin x \cos x) = 1 - \frac{1}{2} \sin(2x)$.Now,differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = 0 - \frac{1}{2} \cos(2x) \cdot 2 = -\cos(2x)$.Calculate $f^{\prime}\left(\frac{\pi}{6}ight)$:$f^{\prime}\left(\frac{\pi}{6}ight) = -\cos\left(2 \cdot \frac{\pi}{6}ight) = -\cos\left(\frac{\pi}{3}ight) = -\frac{1}{2}$.

If $f(x) = \frac{\sin^2 x}{1+\cot x} + \frac{\cos^2 x}{1+\tan x}$,then the value of $f^{\prime}\left(\frac{\pi}{6}\right)$ is equal to

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