Solveeit Logo
Login

Question

Question: \[\frac{\sin^{2}A - \sin^{2}B}{\sin A\cos A - \sin B\cos B} =\]...

sin⁡2A−sin⁡2Bsin⁡Acos⁡A−sin⁡Bcos⁡B=\frac{\sin^{2}A - \sin^{2}B}{\sin A\cos A - \sin B\cos B} =

A

tan⁡(A−B)\tan(A - B)

B

tan⁡(A+B)\tan(A + B)

C

cot⁡(A−B)\cot(A - B)

D

cot⁡(A+B)\cot(A + B)

Answer

tan⁡(A+B)\tan(A + B)

Explanation

Solution

sin⁡2A−sin⁡2Bsin⁡Acos⁡A−sin⁡Bcos⁡B=2sin⁡(A+B)sin⁡(A−B)sin⁡2A−sin⁡2B\frac{\sin^{2}A - \sin^{2}B}{\sin A\cos A - \sin B\cos B} = \frac{2\sin(A + B)\sin(A - B)}{\sin 2A - \sin 2B}

=2sin⁡(A+B)sin⁡(A−B)2cos⁡(A+B)sin⁡(A−B)=tan⁡(A+B)= \frac{2\sin(A + B)\sin(A - B)}{2\cos(A + B)\sin(A - B)} = \tan(A + B).