If frac{sin ^{-1} x}{a}=frac{cos ^{-1} x}{b}= | vedclass

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(C) Given $\frac{\sin ^{-1} x}{a} = \frac{\cos ^{-1} x}{b} = \frac{	an ^{-1} y}{c} = k$ (where $k$ is a constant). Then $\sin ^{-1} x = ak$,$\cos ^{-

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$\frac{1-y^{2}}{1+y^{2}}$

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(C) Given $\frac{\sin ^{-1} x}{a} = \frac{\cos ^{-1} x}{b} = \frac{	an ^{-1} y}{c} = k$ (where $k$ is a constant). Then $\sin ^{-1} x = ak$,$\cos ^{-1} x = bk$,and $	an ^{-1} y = ck$. We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$. Substituting the values,we get $ak + bk = \frac{\pi}{2}$,which implies $k(a + b) = \frac{\pi}{2}$,or $a + b = \frac{\pi}{2k}$. Now,we need to evaluate $\cos \left(\frac{\pi c}{a + b}ight)$. Substituting $a + b = \frac{\pi}{2k}$,we get $\cos \left(\frac{\pi c}{\frac{\pi}{2k}}ight) = \cos (2ck)$. Since $	an ^{-1} y = ck$,we have $\cos (2ck) = \cos (2 	an ^{-1} y)$. Using the formula $\cos (2	heta) = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$,where $	heta = 	an ^{-1} y$,we get $\cos (2 	an ^{-1} y) = \frac{1 - y^2}{1 + y^2}$.

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

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