Question

In a $△ABC,$ if $△=a^2-{(b-c)}^2,$ then $\tan A=$

• A. $\frac{15}{16}$
• B. $\frac{1}{2}$
• C. $\frac{8}{17}$
• D. $\frac{8}{15}$

Answer 1

The correct option is D $\frac{8}{15}$

$△=a^2-{(b-c)}^2$
$=(a+b-c)(a-b+c)$
$=(2s-c-c)(2s-b-b)$
$=(2s-2c)(2s-2b)$
$=4(s-b)(s-c)$
$\Rightarrow \frac{△}{4}=(s-b)(s-c)$
$\Rightarrow \frac{1}{4}=\frac{(s-b)(s-c)}{△}$
$\Rightarrow \frac{1}{4}=\tan \frac{A}{2}$
$\therefore \tan A=\frac{2\tan \frac{A}{2}}{1+{\tan }^2\frac{A}{2}}$
$=\frac{2\times \frac{1}{4}}{1+{\left(\frac{1}{4}\right)}^2}$
$=\frac{8}{15}$ on simplification.