Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. Then, $\left(\cot \frac{A-B}{2}\right)\times \left(\tan \frac{A+B}{2}\right)$ is equal to:

• A. $\frac{a+b}{a-b}$
• B. $\frac{a-b}{a+b}$
• C. $\frac{a(a-b)}{b(a+b)}$
• D. none of these

Answer 1

The correct option is A $\frac{a+b}{a-b}$

According to napiers analogy, we can write

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

Also we know that

$A+B+C=180$

$A+B=180-C$

So,

$\tan \left[\frac{1}{2}(A+B)\right]=\tan [90-C]=\cot \frac{C}{2}...........\left(2\right)$

Putting equation (1) and (2) in the equation given in the question, we get

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

$\tan$ and $\cot$ cancel out and we get

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