In a triangle ABC,if rr1=12 , then 4tanA2tanB2+tanC2=

rr1=12⇒tanB2tanC2=12We have tanA2⋅tanB2+tanB2⋅tanC2+tanC2⋅tanA2=1 tanA2⋅tanB2+tanC2⋅tanA2=12∴ 4tanA2tanB2+tanC2=2

In a triangle ABC,if rr1=12 , then 4tanA2tanB2+tanC2=

rr1=12⇒tanB2tanC2=12We have tanA2⋅tanB2+tanB2⋅tanC2+tanC2⋅tanA2=1 tanA2⋅tanB2+tanC2⋅tanA2=12∴ 4tanA2tanB2+tanC2=2

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In a triangle ABC,if $\frac{r}{r_{1}} = \frac{1}{2}$ , then $4 \tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) =$

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answer is B.

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Detailed Solution

$\frac{r}{r_{1}} = \frac{1}{2} \Rightarrow \tan \frac{B}{2} \tan \frac{C}{2} = \frac{1}{2}$

We have $\tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{B}{2} \cdot \tan \frac{C}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = 1$

$\tan \frac{A}{2} \cdot \tan \frac{B}{2} + \tan \frac{C}{2} \cdot \tan \frac{A}{2} = \frac{1}{2}$
$∴ 4 \tan \frac{A}{2} (\tan \frac{B}{2} + \tan \frac{C}{2}) = 2$