Question

Solve :$\frac{si{n}^{2}A-si{n}^{2}B}{sinAcosA-sinBcosB}=$

• A. $tan(A-B)$
• B. $tan(A+B)$
• C. $cot(A-B)$
• D. $cot(A+B)$

Answer 1

The correct option is B $tan(A+B)$

Solution:- (B) $\tan (A+B)$

$\frac{({\sin }^{2}A-{\sin }^{2}B)}{\sin A\cos A-\sin B\cos B}$

$=\frac{2({\sin }^{2}A-{\sin }^{2}B)}{2(\sin A\cos A-\sin B\cos B)}$

$=\frac{(2{\sin }^{2}A-2{\sin }^{2}B)}{2\sin A\cos A-2\sin B\cos B}$

$=\frac{(1-\cos 2A)-(1-\cos 2B)}{\sin 2A-\sin 2B}$

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

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

$=\tan (A+B)$

Hence $\frac{({\sin }^{2}A-{\sin }^{2}B)}{\sin A\cos A-\sin B\cos B}=\tan (A+B)$.