Question

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$\pink{Question}$

If msinA =nsin(A+2B), prove that tan (A+B)cotB =(m+n)÷(m-n)​

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Answer 1

Answer:

Solution:-

Given :

msinA = nsin(A+2B)

$\begin{gathered} \sf : \longmapsto\green { \frac{m}{n} = \frac{ \sin(A + 2B)}{ \sin A} } \\ \end{gathered}$

Apply Componendo and dividendo

$\begin{gathered}\sf:\longmapsto \red{ \frac{m + n}{m - n} = \frac{ \sin( A + 2B) + \sin A }{ \sin( A+ 2B) - \sin A} } \\ \\ \sf:\longmapsto \blue{ \frac{m + n}{m - n} = \frac{ \sin( A+ B) \cos }{ \cos(A +B ) \sin} } \: \: \: \: \: \: \: \: \: \: \\ \\ \bf:\longmapsto { \frac{m + n}{m - n} = \tan(A + B) \cot B } \: \: \: \end{gathered}$

Hence Proved.

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