Question

In a ${\Delta }^{le}ABC,tan\frac{A}{2},Tan\frac{B}{2}Tan\frac{C}{2}$ are in H.P. then the value of cot $\left(\frac{A}{2}\right).cot\left(\frac{C}{2}\right)=$

• A. 1
• B. 2
• C. 3
• D. $\frac{1}{2}$

Answer 1

The correct option is C

3

$tan\frac{A}{2},Tan\frac{B}{2},Tan\frac{C}{2}inHP\Rightarrow \frac{2}{Tan\frac{B}{2}}=\frac{1}{Tan\frac{A}{2}}+\frac{1}{Tan\frac{C}{2}}\Rightarrow 2Tan\frac{A}{2}Tan\frac{C}{2}=Tan\frac{B}{2}Tan\frac{C}{2}+Tan\frac{B}{2}tan\frac{A}{2}$
$[In\Delta ABC,tan\frac{A}{2}Tan\frac{B}{2}+tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}=1]$
$=2Tan\frac{A}{2}tan\frac{C}{2}=1-tan\frac{A}{2}tan\frac{C}{2}=3Tan\frac{A}{2}tan\frac{C}{2}=1=tan\frac{A}{2}tan\frac{C}{2}=\frac{1}{3}\Rightarrow cot\frac{A}{2}cot\frac{C}{2}=3.$