In A B C, 1- A 2 B 2 = - Vidyadip

MCQ

In $\triangle A B C, 1-\tan \frac{A}{2} \tan \frac{B}{2}=$

✓
$\frac{2 c}{a+b+c}$
B
$\frac{a}{a+b+c}$
C
$\frac{2}{a+b+c}$
D
$\frac{4 a}{a+b+c}$

✓

Answer

Correct option: A.

$\frac{2 c}{a+b+c}$

(A) $1-\tan \frac{ A }{2} \tan \frac{B}{2}=\frac{\cos \frac{ A }{2} \cos \frac{B}{2}-\sin \frac{ A }{2} \sin \frac{B}{2}}{\cos \frac{A}{2} \cos \frac{B}{2}}$
$=\frac{\cos \left(\frac{ A }{2}+\frac{ B }{2}\right)}{\cos \frac{ A }{2} \cos \frac{B}{2}}$
$=\frac{\sin \frac{ C }{2}}{\cos \frac{A}{2} \cos \frac{B}{2}}$
$=\left[\frac{(s-a)(s-b) b c \cdot a c}{a b \cdot s(s-a) s(s-b)}\right]^{1 / 2}$
$=\frac{ c }{ s }=\frac{2 c }{ a + b + c }$

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