Question

$\frac{{\sin }^{2}A-{\sin }^{2}B}{\sin A\cos A-\sin B\cos B}$ is equal to

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

Answer 1

Answer: (B)

$\tan (A+B)$

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

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

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

$=\frac{{\sin }^{2}A+1-{\cos }^{2}A-{\sin }^{2}B-1+{\cos }^{2}B}{\sin 2A-\sin 2B}$ ........ $[\because sin²\theta +cos²\theta =1]$ and $[\because \sin 2x=2\sin x\cos x]$

$=\frac{\cos 2B-\cos 2A}{\sin 2A-\sin 2B}$ ................. $[\because \cos 2\theta =cos²\theta -sin²\theta ]$

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

$=\tan (A+B)$