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

Question

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

${{1 - {y^2}} \over {2y}}$

${{1 - {y^2}} \over {y\sqrt y }}$

$1 - {y^2}$

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

Solution

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