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If \( y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \), then \( \frac{d y}{d x} \) is equal to
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\( 11 \)
Given that, \( y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \)
\( =\tan ^{-1}\left(\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\right)=\tan ^{-1}\left(\frac{\tan x+1}{1-\tan x}\right) \)
\( =\tan ^{-1}\left\{\tan \left(\frac{\Pi}{4}+x\right)\right\}=\frac{I}{4}+x \)
\( \Rightarrow y=\frac{I}{4}+x \)
So, \( \frac{d y}{d x}=0+1=1 \)
\( =\tan ^{-1}\left(\frac{\frac{\sin x}{\cos x}+\frac{\cos x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\right)=\tan ^{-1}\left(\frac{\tan x+1}{1-\tan x}\right) \)
\( =\tan ^{-1}\left\{\tan \left(\frac{\Pi}{4}+x\right)\right\}=\frac{I}{4}+x \)
\( \Rightarrow y=\frac{I}{4}+x \)
So, \( \frac{d y}{d x}=0+1=1 \)
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