In A B C, (A-B) (A+B) = - Vidyadip

MCQ

In $\triangle A B C, \frac{\sin (A-B)}{\sin (A+B)}=$

✓
$\frac{a^2-b^2}{c^2}$
B
$\frac{a^2+b^2}{c^2}$
C
$\frac{c^2}{a^2-b^2}$
D
$\frac{c^2}{a^2+b^2}$

✓

Answer

Correct option: A.

$\frac{a^2-b^2}{c^2}$

(A) $\frac{\sin ( A - B )}{\sin ( A + B )}=\frac{\sin A \cos B -\sin B \cos A }{\sin C }$
$=\frac{ a }{ c } \cos B -\frac{ b }{ c } \cos A$
But $\cos B =\frac{ a ^2+ c ^2- b ^2}{2 ac }, \cos A =\frac{ b ^2+ c ^2- a ^2}{2 bc }$
$\therefore \quad \frac{\sin ( A - B )}{\sin ( A + B )}=\frac{1}{2 c ^2}\left( a ^2+ c ^2- b ^2- b ^2- c ^2+ a ^2\right)$
$=\frac{a^2-b^2}{c^2}$

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