The value of cos (tan^{-1} x) is equal to : . | vedclass

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(A) Let $	heta = 	an^{-1} x$. Then,$	an 	heta = x = \frac{x}{1}$. We know that in a right-angled triangle,$	an 	heta = \frac{	ext{opposite}}{

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $	heta = 	an^{-1} x$. Then,$	an 	heta = x = \frac{x}{1}$. We know that in a right-angled triangle,$	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}}$. Here,the opposite side is $x$ and the adjacent side is $1$. Using the Pythagorean theorem,the hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{1+x^2}$. Now,$\cos 	heta = \frac{	ext{adjacent}}{	ext{hypotenuse}} = \frac{1}{\sqrt{1+x^2}}$. Therefore,$\cos (	an^{-1} x) = \frac{1}{\sqrt{1+x^2}}$. Thus,the correct option is $A$.

The value of $\cos (\tan^{-1} x)$ is equal to : . . . . . . (Where $|x| < 1$)

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