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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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Given, $f(x)=\frac{\sin ^{2} x}{1+\cot x}+\frac{\cos ^{2} x}{1+\tan x}$
$$
\begin{gathered}
f(x)=\frac{\sin ^{2} x}{1+\frac{1}{\tan x}}+\frac{\cos ^{2} x}{1+\tan x} \\
f(x)=\frac{\tan x \cdot \sin ^{2} x+\cos ^{2} x}{(1+\tan x)}=\frac{\sin ^{3} x+\cos ^{3} x}{(\cos x+\sin x)} \\
f(x)=\frac{(\sin x+\cos x)}{(\sin x+\cos x)}\left(\sin ^{2} x+\cos ^{2} x\right. \\
f^{\prime}\left(\frac{\pi}{4}\right)=-\cos 2\left(\frac{\pi}{4}\right)=-\cos \frac{\pi}{2}=0
\end{gathered}
$$
$$
\begin{gathered}
f(x)=\frac{\sin ^{2} x}{1+\frac{1}{\tan x}}+\frac{\cos ^{2} x}{1+\tan x} \\
f(x)=\frac{\tan x \cdot \sin ^{2} x+\cos ^{2} x}{(1+\tan x)}=\frac{\sin ^{3} x+\cos ^{3} x}{(\cos x+\sin x)} \\
f(x)=\frac{(\sin x+\cos x)}{(\sin x+\cos x)}\left(\sin ^{2} x+\cos ^{2} x\right. \\
f^{\prime}\left(\frac{\pi}{4}\right)=-\cos 2\left(\frac{\pi}{4}\right)=-\cos \frac{\pi}{2}=0
\end{gathered}
$$
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