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

Question

(C) Let $	an^{-1} x = y$. Then $	an y = x$.
Since $	an y = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{x}{1}$,we can consider a right-angled t

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $	an^{-1} x = y$. Then $	an y = x$.
Since $	an y = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{x}{1}$,we can consider a right-angled triangle where the opposite side is $x$ and the adjacent side is $1$.
The hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{1+x^2}$.
Therefore,$\sin y = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{x}{\sqrt{1+x^2}}$.
Thus,$\sin(	an^{-1} x) = \sin y = \frac{x}{\sqrt{1+x^2}}$.

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

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