Question

In triangle $ABC,\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}$ are in $H.P$., then the value of $\cot \frac{A}{2}\times \cot \frac{C}{2}$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\to \tan A/2,\tan B/2,\tan C/2$ are in $HP$

$\Rightarrow \frac{1}{\tan A/2},\frac{1}{\tan B/2},\frac{1}{\tan C/2}$ are in $A.P$

$\Rightarrow \frac{2}{\tan B/2}=\frac{1}{\tan A/2}+\frac{1}{\tan C/2}$

$\Rightarrow 2\cot B/2=\cot A/2+\cot C/2$

$\Rightarrow 2\cot \left(\frac{\pi -(A+C)}{2}\right)=\cot A/2+\cot C/2$

$\Rightarrow 2\tan \left(\frac{A+C}{2}\right)=\cot A/2+\cot C/2$

$\Rightarrow 2\left[\frac{\tan A/2+\tan C/2}{1-\tan A/2\tan C/2}\right]=\frac{\tan A/2+\tan C/2}{\tan A/2\tan C/2}$

let $\tan A/2\tan C/2=y$

$\Rightarrow \frac{2}{1-y}=\frac{1}{y}\Rightarrow 2y=1-y$

$\Rightarrow y=-1/3$

$\Rightarrow \cot A/2\cot C/2=\frac{1}{y}=3\Rightarrow \left(C\right)$