cos left(tan^{-1} xright) = . . . . . . . (|x | vedclass

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(C) Let $	heta = 	an^{-1} x$. Then,$	an 	heta = x = \frac{x}{1}$. In a right-angled triangle,the opposite side is $x$ and the adjacent side is $1$

Vedclass · Accepted Answer

$\frac{1}{\sqrt{1+x^2}}$

Vedclass · Answer

(C) Let $	heta = 	an^{-1} x$. Then,$	an 	heta = x = \frac{x}{1}$. In a right-angled triangle,the opposite side is $x$ and the adjacent side is $1$. The hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{1+x^2}$. Therefore,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{1}{\sqrt{1+x^2}}$. Thus,$\cos(	an^{-1} x) = \frac{1}{\sqrt{1+x^2}}$. The correct option is $C$.

$\cos \left(\tan^{-1} x\right) = . . . . . . . (|x| < 1)$.

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