Question

If $\tan \left(A\right)=\frac{1-\cos \left(B\right)}{\sin \left(B\right)}$, then correct relation between $\tan \left(2A\right)\mathrm{and}\tan \left(B\right)$ is

• A. $\tan \left(2A\right)=\tan \left(B\right)$
• B. $\tan \left(2A\right)={\tan }^2\left(B\right)$
• C. $\tan \left(2A\right)=\tan {\left(2\right)}^B+2\tan \left(B\right)$
• D. None of these

Answer 1

The correct option is A

$\tan \left(2A\right)=\tan \left(B\right)$

Explanation for the correct option:

Step 1: Given that,

$\tan \left(A\right)=\frac{1-\cos \left(B\right)}{\sin \left(B\right)}...\left(1\right)$

Use the formulas,

$\cos \left(2\theta \right)=1-2{\sin }^2\left(\theta \right)$ and $\sin \left(2\theta \right)=2\sin \left(\theta \right)\cos \left(\theta \right)$

Step 2: From equation $\left(1\right)$,

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

Hence, option A is correct.