Trigonometric Equations Question 259
Question: In any triangle $ ABC,\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}= $
Options:
A) $ \frac{a-b}{a+b} $
B) $ \frac{a-b}{c} $
C) $ \frac{a-b}{a+b+c} $
D) $ \frac{c}{a+b} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}=\frac{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}-\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}} $ $ =\frac{(s-b)\sqrt{s(s-c)}-(s-a)\sqrt{s(s-c)}}{(s-b)\sqrt{s(s-c)}+(s-a)\sqrt{s(s-c)}} $ $ =\frac{\sqrt{s(s-c)}(s-b-s+a)}{\sqrt{s(s-c)}(s-b+s-a)}=\frac{a-b}{c} $ .
BETA
