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If $\tan \left(\frac{x}{2}\right)=\frac{m}{n}$, then the value of $m$ $\sin (x)+n \cos (x)$ is equal to
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Verified Answer
The correct answer is:
$n$
It is given that $\tan \frac{x}{2}=\frac{m}{n}$, so
$$
\begin{aligned}
m \sin (x)+n \cos (x) & =m\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+n \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \\
& =m \frac{2\left(\frac{m}{n}\right)}{1+\frac{m^2}{n^2}}+n \frac{1-\left(\frac{m^2}{n^2}\right)}{1+\left(\frac{m^2}{n^2}\right)} \\
& =\frac{2 m^2 n}{m^2+n^2}+\frac{n\left(n^2-m^2\right)}{m^2+n^2} \\
& =\frac{n\left(2 m^2+n^2-m^2\right)}{m^2+n^2}=n
\end{aligned}
$$
$$
\begin{aligned}
m \sin (x)+n \cos (x) & =m\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+n \frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}} \\
& =m \frac{2\left(\frac{m}{n}\right)}{1+\frac{m^2}{n^2}}+n \frac{1-\left(\frac{m^2}{n^2}\right)}{1+\left(\frac{m^2}{n^2}\right)} \\
& =\frac{2 m^2 n}{m^2+n^2}+\frac{n\left(n^2-m^2\right)}{m^2+n^2} \\
& =\frac{n\left(2 m^2+n^2-m^2\right)}{m^2+n^2}=n
\end{aligned}
$$
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