sin (tan^{-1} x),where |x|

Question

(A) Let $	an^{-1} x = 	heta$. Then,$	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

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

Vedclass · Answer

(A) Let $	an^{-1} x = 	heta$. Then,$	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 $\sqrt{(	ext{opposite})^2 + (	ext{adjacent})^2} = \sqrt{x^2 + 1^2} = \sqrt{1+x^2}$. Now,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$. Therefore,$\sin (	an^{-1} x) = \frac{x}{\sqrt{1+x^2}}$. The correct option is $A$.

$\sin (\tan^{-1} x)$,where $|x| < 1$,is equal to:

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