If tan

Given:

$\tan \frac{x}{2}=\frac{m}{n}$

$\Rightarrow \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{m}{n}$

Let $\sin \frac{x}{2}$ be $m k$ and $\cos \frac{x}{2}$ be $n k$.

Now,

$m \sin x+n \cos x=2 m \sin \frac{x}{2} \cos \frac{x}{2}+n\left(\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\right)$

$=2 m \times m k \times n k+n\left(n^{2} k^{2}-m^{2} k^{2}\right)$

$=2 m^{2} k^{2} n+n k^{2}\left(n^{2}-m^{2}\right)$

$=n k^{2}\left(2 m^{2}+n^{2}-m^{2}\right)$

$=n k^{2}\left(m^{2}+n^{2}\right)$

$=n\left(m^{2} k^{2}+n^{2} k^{2}\right)$

$=n\left(\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}\right)$

$=n(1)$

$\therefore m \sin x+n \cos x=n$