Question

In $\Delta ABC,$ if $a=(b-c)\sec \theta ,$ then $\frac{2\sqrt{bc}}{|b-c|}\sin \frac{A}{2}=$

• A. $\cot \theta$
• B. $\cos \theta$
• C. $\sec \theta$
• D. $\tan \theta$

Answer 1

The correct option is D $\tan \theta$

Given: $a=(b-c)\sec \theta$
$\Rightarrow \frac{a}{b-c}=\sec \theta$
Now using ${\tan }^2\theta ={\sec }^2\theta -1$, we have:
${\tan }^2\theta =\frac{a^2}{(b-c{)}^2}-1=\frac{a^2-(b-c{)}^2}{(b-c{)}^2}=\frac{(a+b-c)(a-b+c)}{(b-c{)}^2}$
$[\because 2s=a+b+c]\Rightarrow {\tan }^2\theta =\frac{(2s-2c)(2s-2b)}{(b-c{)}^2}$
$=\frac{4(s-b)(s-c)}{(b-c{)}^2}=\frac{4bc}{(b-c{)}^2}\cdot \frac{(s-b)(s-c)}{bc}[\because {\sin }^2\frac{A}{2}=\frac{(s-b)(s-c)}{bc}]$

$\Rightarrow {\tan }^2\theta =\frac{4bc}{(b-c{)}^2}{\sin }^2\frac{A}{2}$
Taking square root of both sides,
$\Rightarrow \frac{2\sqrt{bc}}{|b-c|}\sin \frac{A}{2}=\tan \theta$