A) \[\frac{2c}{a+b+c}\]
B) \[\frac{a}{a+b+c}\]
C) \[\frac{2}{a+b+c}\]
D) \[\frac{4a}{a+b+c}\]
Correct Answer: A
Solution :
\[1-\tan \frac{A}{2}\tan \frac{B}{2}=\frac{\cos \frac{A}{2}\cos \frac{B}{2}-\sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}\] \[=\frac{\cos \left( \frac{A}{2}+\frac{B}{2} \right)}{\cos \frac{A}{2}\cos \frac{B}{2}}=\frac{\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}\] \[={{\left[ \frac{(s-a)(s-b)bc.\,ac}{ab.s(s-a)s(s-b)} \right]}^{1/2}}\]\[=\frac{c}{s}=\frac{2c}{a+b+c}\].
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