If tan left(frac{x}{2}right) = frac{m}{n},the | vedclass

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(D) Given that $	an \left(\frac{x}{2}ight) = \frac{m}{n}$.We use the half-angle formulas: $\sin (x) = \frac{2 	an (x/2)}{1 + 	an^2 (x/2)}$ and $\

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$n$

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(D) Given that $	an \left(\frac{x}{2}ight) = \frac{m}{n}$.We use the half-angle formulas: $\sin (x) = \frac{2 	an (x/2)}{1 + 	an^2 (x/2)}$ and $\cos (x) = \frac{1 - 	an^2 (x/2)}{1 + 	an^2 (x/2)}$.Substituting these into the expression $m \sin (x) + n \cos (x)$:$m \left( \frac{2 	an (x/2)}{1 + 	an^2 (x/2)} ight) + n \left( \frac{1 - 	an^2 (x/2)}{1 + 	an^2 (x/2)} ight)$$= m \left( \frac{2(m/n)}{1 + (m^2/n^2)} ight) + n \left( \frac{1 - (m^2/n^2)}{1 + (m^2/n^2)} ight)$$= m \left( \frac{2m/n}{(n^2 + m^2)/n^2} ight) + n \left( \frac{(n^2 - m^2)/n^2}{(n^2 + m^2)/n^2} ight)$$= \frac{2m^2 n}{n^2 + m^2} + \frac{n(n^2 - m^2)}{n^2 + m^2}$$= \frac{2m^2 n + n^3 - nm^2}{m^2 + n^2}$$= \frac{m^2 n + n^3}{m^2 + n^2} = \frac{n(m^2 + n^2)}{m^2 + n^2} = n$.

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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