A) \[\sqrt{3}\]
B) \[\frac{1}{\sqrt{3}}\]
C) 0
D) \[-\sqrt{3}\]
Correct Answer: C
Solution :
Given, \[f(x)=\frac{{{\sin }^{2}}x}{1+\cot x}+\frac{{{\cos }^{2}}x}{1+\tan x}\] \[f(x)=\frac{{{\sin }^{2}}x}{1+\frac{1}{\tan x}}+\frac{{{\cos }^{2}}x}{1+\tan x}\] \[f(x)=\frac{\tan x.{{\sin }^{2}}x+{{\cos }^{2}}x}{(1+\tan x)}=\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{(\cos x+\sin x)}\] \[f(x)=\frac{(\sin x+\cos x)}{(\sin x+\cos x)}({{\sin }^{2}}x+{{\cos }^{2}}x\] \[-\sin x.\cos x)\] \[f(x)=1-\frac{1}{2}\sin 2x,f'(x)=-\cos 2x\] \[f'\left( \frac{\pi }{4} \right)=-\cos 2\left( \frac{\pi }{4} \right)=-\cos \frac{\pi }{2}=0\]
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