JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 2)
If $${({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2} = a$$; 0 < x < 1, a $$\ne$$ 0, then the value of 2x2 $$-$$ 1 is :
$$\cos \left( {{{4a} \over \pi }} \right)$$
$$\sin \left( {{{2a} \over \pi }} \right)$$
$$\cos \left( {{{2a} \over \pi }} \right)$$
$$\sin \left( {{{4a} \over \pi }} \right)$$
Explanation
Given $$a = {({\sin ^{ - 1}}x)^2} - {({\cos ^{ - 1}}x)^2}$$
$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$
$$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$
$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$
$$ \Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$$
$$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$$
$$ = ({\sin ^{ - 1}}x + {\cos ^{ - 1}}x)({\sin ^{ - 1}}x - {\cos ^{ - 1}}x)$$
$$ = {\pi \over 2}\left( {{\pi \over 2} - 2{{\cos }^{ - 1}}x} \right)$$
$$ \Rightarrow 2{\cos ^{ - 1}}x = {\pi \over 2} - {{2a} \over \pi }$$
$$ \Rightarrow {\cos ^{ - 1}}(2{x^2} - 1) = {\pi \over 2} - {{2a} \over \pi }$$
$$ \Rightarrow 2{x^2} - 1 = \cos \left( {{\pi \over 2} - {{2a} \over \pi }} \right)$$
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