Question

If in triangle ABC, $\angle B={90}^{\circ },$ then ${\tan }^2\left(A/2\right)$ is

• A. $\frac{b-c}{b+c}$
• B. $\frac{b+c}{b-c}$
• C. $\frac{2bc}{b-c}$
• D. none of these

Answer 1

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

Given $\angle B={90}^0$

$\therefore B-C=A$

Thus using above formula, $\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot \frac{A}{2}$

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