Question

Solve ${{\sin }^{-1}}\,2x+{{\cos }^{-1}}\,2x+2\,{{\tan }^{-1}}\,x=\pi$

• A. $1$
• B. $0$
• C. $-1$
• D. $1/\sqrt{2}$

Answer 1

Answer: (A)

$1$

Answer 2

We have, ${{\sin }^{-1}}\,2x+{{\cos }^{-1}}2x+2{{\tan }^{-1}}x=x$ $\Rightarrow$ $\frac{\pi }{2}+2{{\tan }^{-1}}x=\pi \,\left( \because \,\,\,{{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} \right)$ $\Rightarrow$ $2{{\tan }^{-1}}x=\frac{\pi }{1}-\frac{\pi }{2}=\frac{\pi }{2}$ $\Rightarrow$ ${{\tan }^{-1}}x=\frac{\pi }{4}$ $\Rightarrow$ $x=\tan \frac{\pi }{4}$ $\Rightarrow$ $x=1$