Trigonometric Equations Question 216
Question: If in a triangle $ ABC $ , $ \angle C=60^{o}, $ then $ \frac{1}{a+c}+\frac{1}{b+c}= $
[IIT 1975]
Options:
A) $ \frac{1}{a+b+c} $
B) $ \frac{2}{a+b+c} $
C) $ \frac{3}{a+b+c} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \cos C=\frac{\pi }{3}\Rightarrow a^{2}+b^{2}-c^{2}=ab $
Þ $ b^{2}+bc+a^{2}+ac=ab+ac+bc+c^{2} $
Þ $ b(b+c)+a(a+c)=(a+c)(b+c) $ Divide by $ (a+c)(b+c) $ and add 2 on both sides Þ $ 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 $ Þ $ \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c} $ .
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