In any triangle A B C, A 2 - B 2 A 2 + B 2 = - Vidyadip

MCQ

In any triangle $A B C, \frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}=$

A
$\frac{a-b}{a+b}$
✓
$\frac{a-b}{c}$
C
$\frac{a-b}{a+b+c}$
D
$\frac{c}{a+b}$

✓

Answer

Correct option: B.

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

(B) $\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}}$
$=\frac{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}-\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}{\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}}$
$=\frac{(s-b) \sqrt{s(s-c)}-(s-a) \sqrt{s(s-c)}}{(s-b) \sqrt{s(s-c)}+(s-a) \sqrt{s(s-c)}}$
$=\frac{\sqrt{s(s-c)}(s-b-s+a)}{\sqrt{s(s-c)}(s-b+s-a)}$
$=\frac{ a - b }{ c }$

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