Question

Question ▶

TanA +Tan B/TanA -TanB=Sin(A+B)/Sin(A-B)

Answer 1

The answer $\bf \frac{\tan A+\tan B}{\tan A - \tan B} = \frac{\sin [A+B]}{\sin [A-B]}$ is proven..

Given:

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

To Find:

$\frac{\tan A+\tan B}{\tan A - \tan B} = \frac{\sin [A+B]}{\sin [A-B]}$

Solution:
We know that

$\tan A = \frac{\sin A}{\cos A}\\ \\\tan B = \frac{\sin B}{\cos B}$

Putting this in the expression, we get

$\frac{\tan A+\tan B}{\tan A - \tan B}=\frac{\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B} }{\frac{\sin A}{\cos A}-\frac{\sin B}{\cos B} }$

$\frac{\tan A+\tan B}{\tan A - \tan B}= \frac{\frac{\sin A \cos B + \sin B\cos A}{\cos A\cos B} }{\frac{\sin A \cos B - \sin B\cos A}{\cos A\cos B}}$

$\frac{\tan A+\tan B}{\tan A - \tan B}= \frac{\sin A \cos B + \sin B\cos A}{\sin A \cos B - \sin B\cos A}$

$\frac{\tan A+\tan B}{\tan A - \tan B}=\frac{\sin[A+B]}{\sin[A-B]}$

Hence we proved that $\bf\frac{\tan A+\tan B}{\tan A - \tan B} = \frac{\sin [A+B]}{\sin [A-B]}$.

#SPJ2