Question

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

• A. $\frac{b-c}{a}$
• B. $\frac{b-c}{b+c}$
• C. $\frac{b+c}{b-c}$
• D. $\frac{b+c}{a}$

Answer 1

Answer: (B)

$\frac{b-c}{b+c}$

${\tan }^2\frac{A}{2}=\frac{1-\cos A}{1+\cos A}$
$=\frac{1-\frac{c}{b}}{1+\frac{c}{b}}$
$=\frac{b-c}{b+c}$