hi ,im struggling with this question,all help greatly appreciated

this is the proof question ..

sin(A+B) / sin(A-B) = (tanA + tanB) / (tanA - tanB )

Printable View

- Nov 21st 2009, 05:32 AMmike--1988proof (compound angles) trig identities
hi ,im struggling with this question,all help greatly appreciated

this is the proof question ..

sin(A+B) / sin(A-B) = (tanA + tanB) / (tanA - tanB ) - Nov 21st 2009, 06:00 AMGrandad
Hello mike--1988$\displaystyle \frac{\sin(A+B)}{\sin(A-B)}=\frac{\sin A \cos B + \cos A \sin B}{\sin A \cos B - \cos A \sin B}$

Divide top and bottom by $\displaystyle \cos A \cos B$:$\displaystyle =\frac{\dfrac{\sin A}{ \cos A} + \dfrac{\sin B}{\cos B}}{\dfrac{\sin A}{ \cos A} - \dfrac{\sin B}{\cos B}}$Grandad

$\displaystyle =\frac{\tan A + \tan B}{\tan A - \tan B}$