Mathematics · 2021
JEE Main · 27 August 2021, Shift 1 · Q76
If ( sin^-1 x)^2-( cos^-1 x)^2= a; 0<x<1, a ≠ 0, then the value of 2 x^2-1 is:
If $\displaystyle \left(\sin ^{-1} x\right)^2-\left(\cos ^{-1} x\right)^2=\mathrm{a} ; 0<x<1, \mathrm{a} \neq 0$, then the value of $\displaystyle 2 x^2-1$ is :
Official answer
(2)
$\displaystyle \sin \left(\frac{2 \mathrm{a}}{\pi}\right)$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.