Question

If $\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c} ; 0<\,x<\,1,$ then the value of $\cos \left(\frac{\pi c }{ a + b }\right)$ is

• A. $\frac{1-y^{2}}{y \sqrt{y}}$
• B. $1- y ^{2}$
• C. $\frac{1-y^{2}}{1+y^{2}}$
• D. $\frac{1-y^{2}}{2 y}$

Answer 1

Answer: (C)

$\frac{\sin ^{-1} x}{r}=a, \frac{\cos ^{-1} x}{r}=b, \frac{\tan ^{-1} y}{r}=c$
So, $a+b=\frac{\pi}{2 r}$
$\cos \left(\frac{\pi c }{ a + b }\right)=\cos \left(\frac{\pi \tan ^{-1} y }{\frac{\pi}{2 r } r }\right)$
$=\cos \left(2 \tan ^{-1} y \right),$ let $\tan ^{-1} y =\theta$
$=\cos (2 \theta)$
$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{1- y ^{2}}{1+ y ^{2}}$