Question

If ${\sin }^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}=\frac{\pi }{p}+\frac{{\cos }^{-1}x}{q}$, $0<x<1$
Find the value of $p+q$

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The correct option is C 6

Let $x=\cos \theta$. Since $0<x<1$ $\Rightarrow$ $0<\sin \theta <1$ $\Rightarrow$ $\theta \in (0,\frac{\pi }{2})$

$\Rightarrow$ ${\sin }^{-1}\left\{\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right\}={\sin }^{-1}\left\{\frac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{2}\right\}$

$={\sin }^{-1}\left\{\frac{\sqrt{2{\cos }^2\frac{\theta }{2}}+\sqrt{2{\sin }^2\frac{\theta }{2}}}{2}\right\}$

$={\sin }^{-1}\left\{\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\sqrt{2}}\right\}$

$={\sin }^{-1}\left\{\sin (\frac{\pi }{4}+\frac{\theta }{2})\right\}$

$=\frac{\pi }{4}+\frac{\theta }{2}$ $=\frac{\pi }{4}+\frac{{\cos }^{-1}x}{2}$

$[\because \theta ϵ(0,\frac{\pi }{2})\Rightarrow (\frac{\pi }{4}+\frac{\theta }{2})ϵ(\frac{\pi }{4},\frac{\theta }{2})]$

$p+q=4+2=6$