Question

sin(A+B)+sin(A-B) / cos(A+B)+cos(A-B) = tanA​

Answer 1

Answer:

Step-by-step explanation:

sinacosb+cosasinb+sinacosb-cosasinb/cosacosb-sinasinb

2sinacosb/2cosacosb

tan a

Answer 2

Answer:

Since,

sin(x+y) = sinx × cosy + cosx × siny

sin(x-y) = sinx × cosy - cosx × siny

cos(x+y) = cosx × cosy - sinx × siny

cos(x-y) = cosx × cosy + sinx × siny

We have to prove that,

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

L.H.S.

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

$=\frac{\sinA.\cos B+\sin B.\cos A+\sinA.\cos B-\sin B.\cos A}{\cosA.\cos B-\sin A.\sin B+\cosA.\cos B+\sin A.\sin B}$

$=\frac{2\sin A\cos B}{2\cos A\cos B}$

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

$=\tan A$

= R.H.S

Hence, proved....