^2 A - ^2 B A A - B B = - Vidyadip

MCQ

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

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

✓

Answer

Correct option: B.

$\tan (A + B)$

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

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

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

$= \tan \,(A + B)$.

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