%%

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

JEE · Math · previous-year question

  1. A.$$\cos \left( {{{4a} \over \pi }} \right)$$
  2. B.$$\sin \left( {{{2a} \over \pi }} \right)$$correct
  3. C.$$\cos \left( {{{2a} \over \pi }} \right)$$
  4. D.$$\sin \left( {{{4a} \over \pi }} \right)$$

Answer

B. $$\sin \left( {{{2a} \over \pi }} \right)$$

Explanation

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

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions