Question

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

If $\angle B={90}^0$ then $\sqrt{\frac{b-c}{b+c}}$ is equal to?

• A. $\tan \frac{A}{2}$
• B. $2\tan \frac{A}{2}$
• C. $3\tan \frac{A}{2}$
• D. None of these

Answer 1

The correct option is A $\tan \frac{A}{2}$

Given, $\angle B={90}^0\Rightarrow A+C=\frac{\pi }{2}\Rightarrow \frac{A}{2}+\frac{C}{2}=\frac{\pi }{4}$

$\Rightarrow \frac{\tan \frac{A}{2}+\tan \frac{C}{2}}{1-\tan \frac{A}{2}\tan \frac{C}{2}}=1$ ...(1)

Now, by Napier's analogy, $\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot \frac{A}{2}$

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

$=\tan \frac{A}{2}\left[\frac{1-\tan \frac{C}{2}}{1+\tan \frac{C}{2}}\right](\because \frac{B}{2}=\frac{\pi }{4})$

$=\tan \frac{A}{2}.\frac{\tan \frac{A}{2}(1+\tan \frac{C}{2})}{(1+\tan \frac{C}{2})}$ (Using (1))

$={\tan }^2\frac{A}{2}\Rightarrow \sqrt{\frac{b-c}{b+c}}=\tan \frac{A}{2}$