Question

Prove that $\frac{\text{sin}(x+y)}{\text{sin}(x-y)}=\frac{\text{tan}x+\text{tan}y}{\text{tan}x-\text{tan}y}$

Answer 1

$=\frac{\text{sin}(x+y)}{\text{sin}(x-y)}$

$=\frac{\text{sin}x\text{cos}y+\text{cos}x\text{sin}y}{\text{sin}x\text{cos}y-\text{cos}x\text{sin}y}$

$\frac{\text{tan}x+\text{tan}y}{\text{tan}x-\text{tan}y}\left[\text{dividing numerator and denominator by cos}xcosy\right]$= R.H.S.