^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)
$\frac{\sin ^2 A-\sin ^2 B}{\sin A \cos A-\sin B \cos B}$
$=\frac{2 \sin (A+B) \sin (A-B)}{\sin 2 A-\sin 2 B}$
$\ldots\left[\because \sin ^2 A-\sin ^2 B=\sin ( A + B ) \sin ( A - B )\right]$
$=\frac{2 \sin (A+B) \sin (A-B)}{2 \cos (A+B) \sin (A-B)}$
$=\tan ( A + B )$

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