cos left[ 2cos^{-1}frac{1}{5} + sin^{-1}frac{ | vedclass

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(B) We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Given expression: $\cos \left[ \cos^{-1}\frac{1}{5} + \cos^{-1}\frac{1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Given expression: $\cos \left[ \cos^{-1}\frac{1}{5} + \cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5} ight]$.Using the property,this simplifies to: $\cos \left[ \frac{\pi}{2} + \cos^{-1}\frac{1}{5} ight]$.Since $\cos(\frac{\pi}{2} + 	heta) = -\sin	heta$,we have: $-\sin \left( \cos^{-1}\frac{1}{5} ight)$.Let $\cos^{-1}\frac{1}{5} = \alpha$,then $\cos\alpha = \frac{1}{5}$.Then $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (\frac{1}{5})^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$.Thus,the expression equals $-\frac{2\sqrt{6}}{5}$.

$\cos \left[ 2\cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5} \right] = $

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