JEE Math Practice 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 :
- A.$${{1 - {y^2}} \over {2y}}$$
- B.$${{1 - {y^2}} \over {y\sqrt y }}$$
- C.$$1 - {y^2}$$
- D.$${{1 - {y^2}} \over {1 + {y^2}}}$$
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