Question

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

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

Answer 1

Answer: (B)

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