Trigonometric Equations Question 191
Question: In $ \Delta ABC,\frac{\sin (A-B)}{\sin (A+B)}= $
[MP PET 1986]
Options:
A) $ \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}} $
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Answer:
Correct Answer: A
Solution:
- $ \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}}{2ac},\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} $
$ \Rightarrow \frac{a}{c}\cos B-\frac{b}{c}\cos A=\frac{1}{2c^{2}} $ $ (a^{2}+c^{2}-b^{2}-b^{2}-c^{2}+a^{2}) $ $ =\frac{a^{2}-b^{2}}{c^{2}} $ .
BETA
