Question:
If $\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c} ; 0
Correct Option: , 2
Solution:
$\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c}$
$\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\sin ^{-1} x+\cos ^{-1} x}{a+b}=\frac{\pi}{2(a+b)}$
Now, $\frac{\tan ^{-1} y}{c}=\frac{\pi}{2(a+b)}$
$2 \tan ^{-1} y=\frac{\pi c}{a+b}$
$\Rightarrow \quad \cos \left(\frac{\pi c}{a+b}\right)=\cos \left(2 \tan ^{-1} y\right)=\frac{1-y^{2}}{1+y^{2}}$
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