Question

If $\tan \left(\frac{A}{2}\right)=\frac{3}{2}$, then $\frac{1+\cos \left(A\right)}{1-\cos \left(A\right)}=$

• A. $-5$
• B. $5$
• C. $\frac{9}{4}$
• D. $\frac{4}{9}$

Answer 1

The correct option is D

$\frac{4}{9}$

Explanation for the correct option:

Step 1: Given that,

$\therefore \tan \left(\frac{A}{2}\right)=\frac{3}{2}$

We know that, $\cos \left(2\theta \right)=2{\cos }^2\left(\theta \right)-1$

Step 2: Use this formula $\cos \left(2\theta \right)=1-2{\sin }^2\left(\theta \right)$

$\begin{array}{rcl}& \therefore & \frac{1+\cos \left(A\right)}{1-\cos \left(A\right)}\\ & =& \frac{2{\cos }^2\left({\displaystyle \frac{A}{2}}\right)}{2{\sin }^2\left({\displaystyle \frac{A}{2}}\right)}\\ & =& co{t}^2\left(\frac{A}{2}\right)\\ & =& \frac{1}{{\tan }^2\left({\displaystyle \frac{A}{2}}\right)}\\ & =& {\left(\frac{2}{3}\right)}^2\\ & =& \left(\frac{4}{9}\right)\end{array}$

Therefore, the value of $\frac{1+\cos \left(A\right)}{1-\cos \left(A\right)}=\frac{4}{9}$

Hence, the correct option is (D).