Question

In a triangle $ABC$, if $c=(a-b)sec\theta$, then $\tan \theta$ is equal to

• A. $2\frac{\sqrt{ab}}{a-b}\sin \left(\frac{C}{2}\right)$
• B. $2\frac{\sqrt{ab}}{a-b}\tan \left(\frac{C}{2}\right)$
• C. $2\frac{\sqrt{ab}}{a-b}sec\left(\frac{C}{2}\right)$
• D. $2\frac{\sqrt{ab}}{a-b}\cos \left(\frac{C}{2}\right)$

Answer 1

The correct option is A

$2\frac{\sqrt{ab}}{a-b}\sin \left(\frac{C}{2}\right)$

Determine the value of $\tan \theta$

Given, $c=(a-b)sec\theta$

$\Rightarrow sec\theta =\frac{c}{a-b}...\left(1\right)$

We know that, ${\tan }^2\theta =se{c}^2\theta –1$, therefore substituting the value of $\left(1\right)$ we have,

$$\begin{gathered} \Rightarrow {\tan }^2\theta ={\left(\frac{c}{a-b}\right)}^2–1 \\ \Rightarrow =\frac{c^2}{{(a-b)}^2}-1 \\ \Rightarrow =\frac{c^2-{(a-b)}^2}{{(a-b)}^2} \\ \Rightarrow =\frac{(c-a+b)\times (c+a-b)}{{(a-b)}^2} \\ \Rightarrow =\frac{(2s-2a)(2s-2b)}{{(a-b)}^2}[\therefore s=\frac{a+b+c}{2}] \\ \Rightarrow =\frac{4\times (s-a)(s-b)}{{(a-b)}^2} \\ \Rightarrow \tan \theta =2\times \frac{\sqrt{(s-a)(s-b)}}{(a-b)} \\ \Rightarrow \tan \theta =2\times \frac{\sqrt{ab}}{a-b}\sin \left(\frac{C}{2}\right)[\because \sin \left(\frac{C}{2}\right)=\sqrt{\frac{(s-a)(s-b)}{ab}}] \end{gathered}$$

Hence, option (A) is correct.