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

Question

Prove that:

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

Proof

L.H.S.

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

Using the identities:

\[ \sin C+\sin D = 2\sin\frac{C+D}{2}\cos\frac{C-D}{2} \]

and

\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]

we get:

\[ = \frac{ 2\sin\frac{(A+B)+(A-B)}{2} \cos\frac{(A+B)-(A-B)}{2} } { 2\cos\frac{(A+B)+(A-B)}{2} \cos\frac{(A+B)-(A-B)}{2} } \]

\[ = \frac{ 2\sin\frac{2A}{2}\cos\frac{2B}{2} } { 2\cos\frac{2A}{2}\cos\frac{2B}{2} } \]

\[ = \frac{ 2\sin A\cos B } { 2\cos A\cos B } \]

Cancelling \[ 2\cos B \] from numerator and denominator:

\[ = \frac{\sin A}{\cos A} \]

\[ = \tan A \]

R.H.S.

\[ = \tan A \]

Hence,

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

Hence proved.