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)

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