Q92 / 180 · KCET 2017 · Mathematics

If \( y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \), then \( \frac{d y}{d x} \) is equal to

  • A
    \( 1 / 2 \)
  • B
    $\pi / 4$
  • C
    \( 00 \)
  • Correct answer
    D
    \( 11 \)

Solution

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, \( \frac{d y}{d x}=0+1=1 \)

Hence, the correct answer is option: 'D'.

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