Trigonometric Equations Question 203
Question: In $ \Delta ABC, $ $ 1-\tan \frac{A}{2}\tan \frac{B}{2}= $
[Roorkee 1973]
Options:
A) $ \frac{2c}{a+b+c} $
B) $ \frac{a}{a+b+c} $
C) $ \frac{2}{a+b+c} $
D) $ \frac{4a}{a+b+c} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ 1-\tan \frac{A}{2}\tan \frac{B}{2}=\frac{\cos \frac{A}{2}\cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}} $ $ =\frac{\cos ( \frac{A}{2}+\frac{B}{2} )}{\cos \frac{A}{2}\cos \frac{B}{2}}=\frac{\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}} $ $ ={{[ \frac{(s-a)(s-b)bc.,ac}{ab.s(s-a)s(s-b)} ]}^{1/2}} $ $ =\frac{c}{s}=\frac{2c}{a+b+c} $ .
BETA
