MCQ
$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=$
  • A
    $\frac{\pi}{2}-x$
  • $\frac{\pi}{4}-x$
  • C
    $\frac{\pi}{2}$
  • D
    $\frac{\pi}{4}$

Answer

Correct option: B.
$\frac{\pi}{4}-x$
(B) $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$
$=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-x\right)\right)$
$=\frac{\pi}{4}-x$

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