Question

Value of ${\tan }^{-1}\left\{\begin{array}{c}\frac{\sin 2-1}{\cos 2}\end{array}\right\}$ is

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

Answer 1

The correct option is B $1-\frac{\pi }{4}$

$ta{n}^{-1}\left\{\begin{array}{c}\frac{\sin 2-1}{\cos 2}\end{array}\right\}$
$ta{n}^{-1}\left\{\begin{array}{c}\frac{2\sin 1cos1-[si{n}^{2}1+{\cos }^{2}1]}{{\cos }^{2}1-si{n}^{2}1}\end{array}\right\}$
$ta{n}^{-1}\left\{\begin{array}{c}\frac{-[\cos 1-\sin 1{]}^2}{(\cos 1-\sin 1)(cos1+sin1)}\end{array}\right\}$
$ta{n}^{-1}\left\{\begin{array}{c}\frac{\cos 1-\sin 1}{\cos 1+\sin 1}\end{array}\right\}$ divide by $cos1=ta{n}^{-1}\left(tan\right(1-\frac{\pi }{4}\left)\right)=1-\frac{\pi }{4}$