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

Question :-

If msinA = nsin(A+2B), then prove that tan(A+B) cotB = $\rm{ \frac{m + n}{m - n}}$

Solution :-

Given : msinA =nsin(A+2B)

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

Apply Componendo and dividendo

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

Hence Proved !

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:msinA = nsin(A + 2B)$

can be rewritten as

$\rm :\longmapsto\:\dfrac{m}{n} = \dfrac{sin(A + 2B)}{sinA}$

We know,

The process of Componendo and Dividendo is defined as

$\red{ \boxed{ \sf{ \: \frac{a}{b} = \frac{c}{d} \: \bf\implies \: \frac{a + b}{a - b} = \frac{c + d}{c - d}}}}$

So,

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{m + n}{m - n} = \dfrac{sin(A + 2B) + sinA}{sin(A + 2B) - sinA}$

We know that,

$\red{ \boxed{ \sf{ \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

and

$\red{ \boxed{ \sf{ \:sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

So, on applying these results, on RHS, we get

$\rm :\longmapsto\:\dfrac{m + n}{m - n} = \dfrac{2sin\bigg[\dfrac{A + 2B + A}{2} \bigg]cos\bigg[\dfrac{A + 2B - A}{2} \bigg]}{2cos\bigg[\dfrac{A + 2B + A}{2} \bigg]sin\bigg[\dfrac{A + 2B - A}{2} \bigg]}$

$\rm :\longmapsto\:\dfrac{m + n}{m - n} = \dfrac{2sin\bigg[\dfrac{2A + 2B}{2} \bigg]cos\bigg[\dfrac{2B}{2} \bigg]}{2cos\bigg[\dfrac{2A + 2B}{2} \bigg]sin\bigg[\dfrac{2B}{2} \bigg]}$

$\rm :\longmapsto\:\dfrac{m + n}{m - n} = \dfrac{sin(A + B)cosB}{cos(A + B)sinB}$

$\rm :\longmapsto\:\dfrac{m + n}{m - n} = \dfrac{tan(A + B)}{tanB}$

$\bf :\longmapsto\:\dfrac{m + n}{m - n} = tan(A + B){cotB}$

Hence,

$\bf :\longmapsto\:\red{ \boxed{ \sf{ \:\dfrac{m + n}{m - n} = tan(A + B){cotB}}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:cosx - cosy = - \: 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}}$

$\red{ \boxed{ \sf{ \:2sinxcosy = sin(x + y) + sin(x - y)}}}$

$\red{ \boxed{ \sf{ \:2cosxcosy = cos(x + y) + cos(x - y)}}}$

$\red{ \boxed{ \sf{ \:2sinxsiny = cos(x - y) - cos(x + y)}}}$