If (cos ^{-1} x)^2-(sin ^{-1} x)^2 0,then | vedclass

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(D) Given the inequality $(\cos ^{-1} x)^2 - (\sin ^{-1} x)^2 > 0$. Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we get: $(\cos ^{-1} x -

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$-1 \leqslant x < \frac{1}{\sqrt{2}}$

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(D) Given the inequality $(\cos ^{-1} x)^2 - (\sin ^{-1} x)^2 > 0$. Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we get: $(\cos ^{-1} x - \sin ^{-1} x)(\cos ^{-1} x + \sin ^{-1} x) > 0$. We know that $\cos ^{-1} x + \sin ^{-1} x = \frac{\pi}{2}$ for $x \in [-1, 1]$. Substituting this,we have: $(\cos ^{-1} x - \sin ^{-1} x) \cdot \frac{\pi}{2} > 0$. Since $\frac{\pi}{2} > 0$,we can divide by it: $\cos ^{-1} x - \sin ^{-1} x > 0$ $\cos ^{-1} x > \sin ^{-1} x$. Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we substitute: $\frac{\pi}{2} - \sin ^{-1} x > \sin ^{-1} x$ $\frac{\pi}{2} > 2 \sin ^{-1} x$ $\sin ^{-1} x Since $\sin 	heta$ is an increasing function,we take the sine of both sides: $x $x Also,the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $x \in [-1, 1]$. Therefore,the solution is $-1 \leqslant x < \frac{1}{\sqrt{2}}$.

If $(\cos ^{-1} x)^2-(\sin ^{-1} x)^2 > 0$,then

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