Prove that: + - = +B -B - Vidyadip

Question

Prove that: $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac {\sin\text{A+B}}{\sin\text{A-B}}$

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Answer

$\text{L.H.S}=$ $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B} }$
$=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\sin\text{A}}{\cos\text{A}}-\frac{\sin\text{B}}{\cos\text{B}}}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$
$=\frac{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}}{{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}}{}}$
$=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}$ $\begin{bmatrix}\because\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\\\text{and,}\sin\text{(A-B)}=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B} \end{bmatrix}$
$=\frac{\sin\text{(A+B)}}{\sin\text{(A-B)}}$
$\therefore\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac{\sin\text{(A+B)}}{\sin\text{(A-B)}} $
$=\text{R.H.S}$
$\text{L.H.S}=\text{R.H.S}$
Hence proved.

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