cos left(2 left(tan ^{-1} frac{1}{5}+tan ^{-1 | vedclass

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(D) We know that $	an ^{-1} x + 	an ^{-1} \frac{1}{x} = \frac{\pi}{2}$ for $x > 0$. Given expression is $\cos \left(2 \left(	an ^{-1} \frac{1}{5}+\

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$-1$

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(D) We know that $	an ^{-1} x + 	an ^{-1} \frac{1}{x} = \frac{\pi}{2}$ for $x > 0$. Given expression is $\cos \left(2 \left(	an ^{-1} \frac{1}{5}+	an ^{-1} 5ight)ight)$. Since $5 > 0$,we have $	an ^{-1} \frac{1}{5} + 	an ^{-1} 5 = \frac{\pi}{2}$. Substituting this into the expression: $\cos \left(2 	imes \frac{\pi}{2}ight) = \cos(\pi)$. We know that $\cos(\pi) = -1$. Therefore,the correct option is $D$.

$\cos \left(2 \left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} 5\right)\right) = $ . . . . . .

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