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if n Tan theta/2=m prove that m sin theta +n cos theta =n

We have, n \tan \frac{θ}{2} = m \tan \frac{θ}{2} = \frac{m}{n} LHS m \sin θ + n \cos θ = m \times \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} + n \times \frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} = m (\frac{2 (\frac{m}{n})}{1 + {(\frac{m}{n})}^{2}}) + n (\frac{1 - {(\frac{m}{n})}^{2}}{1 + {(\frac{m}{n})}^{2}}) = \frac{\frac{2 m^{2}}{n}}{\frac{n^{2} + m^{2}}{n^{2}}} + \frac{\frac{n^{2} - m^{2}}{n}}{\frac{n^{2} + m^{2}}{n^{2}}} = \frac{2 m^{2} n}{m^{2} + n^{2}} + \frac{n (n^{2} - m^{2})}{m^{2} + n^{2}} = \frac{n (2 m^{2} + n^{2} - m^{2})}{m^{2} + n^{2}} = \frac{n (m^{2} + n^{2})}{m^{2} + n^{2}} = n = RHS Hence Proved .

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