Question

In a $△ABC,$ $a,b,c$ are the sides of the triangle opposite to the angles $A,B,C$ respectively. If $b+c=3a,$ then $\cot \frac{B}{2}\cdot \cot \frac{C}{2}$ has the value equal to

• A. 2
• B. 2.00
• C. 2.0
• D. 02

Answer 1

Answer: (A), (B), (C), (D)

$\cot \frac{B}{2}\cdot \cot \frac{C}{2}=\frac{s(s-b)}{\Delta }\cdot \frac{s(s-c)}{\Delta }=\frac{s(s-b)}{\Delta }\cdot \frac{s(s-c)}{\Delta }\cdot \frac{s-a}{s-a}=\frac{s\cdot {△}^2}{{△}^2\cdot (s-a)}=\frac{s}{s-a}=\frac{2s}{2s-2a}$
But given that $a+b+c=4a\Rightarrow 2s=4a$
Hence, $\cot \frac{B}{2}\cdot \cot \frac{C}{2}=\frac{4a}{2a}=2$