Q73 / 160 · WBJEE 2010 · Mathematics

Value of $\tan ^{-1}\left(\frac{\sin 2-1}{\cos 2}\right)$ is

  • A
    $\frac{\pi}{2}-1$
  • Correct answer
    B
    $1-\frac{\pi}{4}$
  • C
    $2-\frac{\pi}{2}$
  • D
    $\frac{\pi}{4}-1$

Solution

Hints : $\tan ^{-1}\left(\frac{\sin 2-1}{\cos 2}\right)=\tan ^{-1}\left(\frac{-(\sin 1-\cos 1)^2}{(\cos 1-\sin 1)(\cos 1+\sin 1)}\right) \quad=-\tan ^{-1}\left(\frac{\cos 1-\sin 1}{\cos 1+\sin 1}\right)=1-\frac{\pi}{4}$

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

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