The value of cos left(2 cos ^{-1} x+sin ^{-1} | vedclass

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(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ We can rewrite this as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)\r

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$-\frac{2 \sqrt{6}}{5}$

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(C) Given expression: $\cos \left(2 \cos ^{-1} x+\sin ^{-1} xight)$ We can rewrite this as: $\cos \left(\cos ^{-1} x + (\cos ^{-1} x+\sin ^{-1} x)ight)$ Using the property $\cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}$,the expression becomes: $\cos \left(\cos ^{-1} x+\frac{\pi}{2}ight)$ Using the trigonometric identity $\cos \left(	heta+\frac{\pi}{2}ight) = -\sin 	heta$: $= -\sin \left(\cos ^{-1} xight)$ Since $\cos ^{-1} x = \sin ^{-1} \sqrt{1-x^2}$,we have: $= -\sin \left(\sin ^{-1} \sqrt{1-x^2}ight) = -\sqrt{1-x^2}$ Substituting $x=\frac{1}{5}$: $= -\sqrt{1-\left(\frac{1}{5}ight)^2} = -\sqrt{1-\frac{1}{25}} = -\sqrt{\frac{24}{25}} = -\frac{2 \sqrt{6}}{5}$

The value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ at $x=\frac{1}{5}$,where $0 \leq \cos ^{-1} x \leq \pi$ and $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}$,is

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