If tan A = x tan B, Prove that sin(A−B)/sin(A+B) = (x−1)/(x+1)

Question

If

\[ \tan A=x\tan B \]

prove that:

\[ \frac{\sin(A-B)}{\sin(A+B)} = \frac{x-1}{x+1} \]

Proof

Given,

\[ \tan A=x\tan B \]

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

\[ \sin A\cos B = x\cos A\sin B \]

Now,

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

\[ = x\cos A\sin B-\cos A\sin B \]

\[ = (x-1)\cos A\sin B \]

Also,

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

\[ = x\cos A\sin B+\cos A\sin B \]

\[ = (x+1)\cos A\sin B \]

Therefore,

\[ \frac{\sin(A-B)}{\sin(A+B)} = \frac{(x-1)\cos A\sin B} {(x+1)\cos A\sin B} \]

\[ = \frac{x-1}{x+1} \]

Hence proved.