If [( ^ - 1x)^2 - ( ^ - 1x)^2 = a]; 0

Question

If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$;  0 < x < 1, a $$\ne$$ 0, then the value of 2x2 - 1 is :

Accepted Answer

Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$

$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$

$$ = {\pi  \over 2}\left( {{\pi  \over 2} - 2{{\cos }^{ - 1}}x} \right)$$

$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi  \over 2} - {{2a} \over \pi }$$

$$ \Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi  \over 2} - {{2a} \over \pi }$$

$$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi  \over 2} - {{2a} \over \pi }} \right)$$

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