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Question

# Write the given expression as an algebraic expression in x.$\sin \left( 2 \tan ^ { -1 } x \right)$

Solution

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Step 1
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Let $\tan^{-1} x= \theta$ then we obtain

$x= \tan \theta$

Therefore

$\sin ( 2 \tan^{-1} x) = \sin (2 \theta)$

According to the double angle formula

$\color{#c34632}{\sin 2x= 2 \sin x \cos x}$

we obtain

$\sin ( 2 \tan^{-1} x) = 2 \sin \theta \cos \theta$

Since $\tan \theta=x$, we obtain

\begin{align*} \sin \theta &= \frac {x}{\sqrt {1+x^2}}\\ \cos \theta &= \frac {1}{\sqrt {1+x^2}} \end{align*}

By substituting $\sin \theta$ and $\cos \theta$ we obtain

\begin{align*} \sin ( 2 \tan^{-1} x) &= 2 \frac {x}{\sqrt {1+x^2}} \cdot \frac {1}{\sqrt {1+x^2}}\\ &= \frac {2x}{1+x^2} \end{align*}

The solution is

$\sin ( 2 \tan^{-1} x) = \frac {2x}{1+x^2}$

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