If (sin ^{-1} x)^{2}-(cos ^{-1} x)^{2}=a ; 0 | vedclass

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(B) Given $a = (\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}$.Using the identity $A^2 - B^2 = (A+B)(A-B)$,we get:$a = (\sin ^{-1} x + \cos ^{-1} x)(\sin ^{-1}

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$\sin \left(\frac{2 a}{\pi}\right)$

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(B) Given $a = (\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}$.Using the identity $A^2 - B^2 = (A+B)(A-B)$,we get:$a = (\sin ^{-1} x + \cos ^{-1} x)(\sin ^{-1} x - \cos ^{-1} x)$.Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have:$a = \frac{\pi}{2}(\sin ^{-1} x - \cos ^{-1} x)$.Substitute $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$:$a = \frac{\pi}{2}(\frac{\pi}{2} - 2 \cos ^{-1} x)$.$\frac{2a}{\pi} = \frac{\pi}{2} - 2 \cos ^{-1} x$.$2 \cos ^{-1} x = \frac{\pi}{2} - \frac{2a}{\pi}$.Taking cosine on both sides:$\cos(2 \cos ^{-1} x) = \cos(\frac{\pi}{2} - \frac{2a}{\pi})$.Using the identity $\cos(2 	heta) = 2 \cos^2 	heta - 1$ and $\cos(\frac{\pi}{2} - 	heta) = \sin 	heta$:$2 x^2 - 1 = \sin(\frac{2a}{\pi})$.

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

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