MCQ
Write the function in the simplest form: $\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
  • A
    $-\frac{\pi}{4}+x$
  • B
    $-\frac{\pi}{4}-x$
  • $\frac{\pi}{4}-x$
  • D
    $\frac{\pi}{4}+x$

Answer

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

$=\tan ^{-1}\left(\frac{1-\left(\frac{\sin x}{\cos x}\right)}{1+\left(\frac{\sin x}{\cos x}\right)}\right)$

$=\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$

$=\tan ^{-1}(1)-\tan ^{-1}(\tan x)$ $\left[\because \frac{-y}{x y}=\tan ^{-1} x-\tan ^{-1} y\right]$

$=\frac{\pi}{4}-x$

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