Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $AC$ is double of $AB$, then the value of $\cot \frac{A}{2}\cot \frac{B-C}{2}$ is equal to?

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

Answer 1

The correct option is C $3$

According to Napier's analogy, we can write

$\cot \frac{1}{2}(B-C)=\frac{b+c}{b-c}\tan \left[\frac{A}{2}\right].........\left(1\right)$

Putting it in equation given in question, we get

$\cot \left[\frac{A}{2}\right]\cot \left[\frac{1}{2}(B-C)\right]=\cot \left[\frac{A}{2}\right]\frac{b+c}{b-c}\tan \left[\frac{A}{2}\right]$

$\cot \left[\frac{A}{2}\right]$ and $\tan \left[\frac{A}{2}\right]$ cancel each other and we get

$\cot \left[\frac{A}{2}\right]\cot \left[\frac{1}{2}(B-C)\right]=\frac{b+c}{b-c}............\left(2\right)$

Now according to the question $AC=2AB$ i.e $b=2c$

Putting this relation in equation $\left(2\right)$, we get

$\cot \left[\frac{A}{2}\right]\cot \left[\frac{1}{2}(B-C)\right]=\frac{2c+c}{2c-c}$

$\cot \left[\frac{A}{2}\right]\cot \left[\frac{1}{2}(B-C)\right]=\frac{3c}{c}$

$\cot \left[\frac{A}{2}\right]\cot \left[\frac{1}{2}(B-C)\right]=3$