Question:
If $\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c} ; 0
Correct Option: , 3
Solution:
$\frac{\sin ^{-1} x}{r}=a, \frac{\cos ^{-1} x}{r}=b, \frac{\tan ^{-1} y}{r}=c$
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 $\tan ^{-1} y=\theta$
$=\cos (2 \theta)$
$=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\frac{1-y^{2}}{1+y^{2}}$
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