Question

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

If $c=(a-b)\sec \alpha$ then $\tan \alpha$ is equal to?

• A. $\frac{2\sqrt{ab}}{a-b}\cos \frac{C}{2}$
• B. $\frac{2\sqrt{ab}}{a-b}\sin \frac{C}{2}$
• C. $\frac{2\sqrt{ab}}{a-b}\tan \frac{C}{2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{2\sqrt{ab}}{a-b}\sin \frac{C}{2}$

We have, $\sec \alpha =\frac{c}{a-b}$

$\Rightarrow {\tan }^2\alpha ={\sec }^2\alpha -1=\frac{c^2}{{(a-b)}^2}-1=\frac{c^2-{(a-b)}^2}{{(a-b)}^2}$

$=\frac{(c-a+b)(2s-2b)}{{(a-b)}^2}=\frac{(2s-2a)(2s-2b)}{{(a-b)}^2}$

$\therefore \tan \alpha =\frac{2\sqrt{(s-a)(s-b)}}{a-b}=\frac{2\sqrt{ab}\sin \frac{C}{2}}{a-b}$