In $\triangle A B C,\left(r_2+r_3\right) \cot \left(\frac{B+C}{2}\right)=$
In $\triangle A B C,\left(r_2+r_3\right) \cot \left(\frac{B+C}{2}\right)=$
- a+b+c
- a
- b
- c
Solution
Given, in $\triangle A B C$
$
\begin{aligned}
& \left(r_2+r_3\right) \cot \left(\frac{B+c}{2}\right)=\left(\frac{\Delta}{s-b}+\frac{\Delta}{s-c}\right) \tan \frac{A}{2} \\
& =\frac{\Delta(s-c+s-b)}{(s-b)(s-c)} \frac{\Delta}{s(s-a)}
\end{aligned}
$
$
=2 s-b-c=a
$
Hence, option (b) is correct
Asked in: AP EAMCET 2019 (20 Apr Shift 2)
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