Question

msinB = nsin(2A+B) then (m+n) tanA=
(m-n) tan (A+B)​

Answer 1

$\textbf{Given:}$

$m\;sinB=n\;sin(2A+B)$

$\textbf{To prove:}$

$(m+n)\,tanA=(m-n)\,tan(A+B)$

$\textbf{Solution:}$

$\text{Consider,}$

$m\;sinB=n\;sin(2A+B)$

$\text{This can be written as}$

$\dfrac{sin(2A+B)}{sinB}=\dfrac{m}{n}$ ......(1)

$\text{Add 1 on bothsides of (1), we get}$

$\dfrac{sin(2A+B)}{sinB}+1=\dfrac{m}{n}+1$

$\dfrac{sin(2A+B)+sinB}{sinB}=\dfrac{m+n}{n}$ .......(2)

$\text{Subtract 1 on bothsides of (1), we get}$

$\dfrac{sin(2A+B)}{sinB}-1=\dfrac{m}{n}-1$

$\dfrac{sin(2A+B)-sinB}{sinB}=\dfrac{m-n}{n}$ .......(3)

$\text{Divide (2) by (3), we get}$

$\dfrac{\frac{sin(2A+B)+sinB}{sinB}}{\frac{sin(2A+B)-sinB}{sinB}}=\dfrac{\frac{m+n}{n}}{\frac{m-n}{n}}$

$\dfrac{sin(2A+B)+sinB}{sin(2A+B)-sinB}=\dfrac{m+n}{m-n}$

$\text{Using the identity}$

$\boxed{sinC+sinD=2\:sin(\frac{C+D}{2})\:cos(\frac{C-D}{2})}$

$\boxed{sinC-sinD=2\:cos(\frac{C+D}{2})\:sin(\frac{C-D}{2})}$

$\dfrac{2\:sin(\frac{2A+B+B}{2})\:cos(\frac{2A+B-B}{2})}{2\:cos(\frac{2A+B+B}{2})\:sin(\frac{2A+B-B}{2})}=\dfrac{m+n}{m-n}$

$\dfrac{sin(A+B)\:cosA}{cos(A+B)\:sinA}=\dfrac{m+n}{m-n}$

$\implies(m-n)\dfrac{sin(A+B)}{cos(A+B)}=(m+n)\dfrac{sinA}{cosA}$

$\implies\bf(m-n)\,tan(A+B)=(m+n)\,tanA$

Answer 2

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