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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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