#### Answer

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#### Work Step by Step

Formula of $\sin 2x$ is $\sin 2x=2\sin x\cos x$.
It is known that, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$.
Divide both sides by ${{\sin }^{2}}x+{{\cos }^{2}}x$ in the above equation:
$\begin{align}
& \frac{\sin 2x}{{{\sin }^{2}}x+{{\cos }^{2}}x}=\frac{2\sin x\cos x}{{{\sin }^{2}}x+{{\cos }^{2}}x} \\
& \frac{\sin 2x}{1}=\frac{2\sin x\cos x}{{{\sin }^{2}}x+{{\cos }^{2}}x}
\end{align}$
Divideby ${{\cos }^{2}}x$ in nominator as well as denominator and get,
$\begin{align}
& \sin 2x=\frac{\left( \frac{2\sin x\cos x}{{{\cos }^{2}}x} \right)}{\left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\cos }^{2}}x} \right)} \\
& =2\frac{\left( \frac{2\sin x}{\cos x} \right)}{\left( 1+\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x} \right)} \\
& =\frac{2\tan x}{1+{{\tan }^{2}}x}
\end{align}$