Prove that: (tan A + tan B)/(tan A − tan B) = sin(A+B)/sin(A−B)

Question

Prove that:

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

Proof

Consider the left-hand side:

\[ \frac{\tan A+\tan B}{\tan A-\tan B} \]

Using

\[ \tan \theta=\frac{\sin\theta}{\cos\theta} \]

we get:

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

Taking LCM in numerator and denominator:

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

Cancelling \[ \cos A\cos B \] from numerator and denominator:

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

Using the identities:

\[ \sin(A+B)=\sin A\cos B+\cos A\sin B \]

and

\[ \sin(A-B)=\sin A\cos B-\cos A\sin B \]

Therefore,

\[ = \frac{\sin(A+B)}{\sin(A-B)} \]

Hence,

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

Hence proved.