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