Question

If $\tan \left(A+B\right)=p$and $\tan \left(A-B\right)=q$, then the value of $\tan \left(2A\right)$ in terms of $p$and $q$is:

• A. $\frac{p+q}{p-q}$
• B. $\frac{p-q}{1+pq}$
• C. $\frac{p+q}{1-pq}$
• D. $\frac{1+pq}{1-p}$

Answer 1

The correct option is C

$\frac{p+q}{1-pq}$

Explanation for the correct option

Given that $\tan \left(A+B\right)=p$and $\tan \left(A-B\right)=q$

$2A=\left(A+B\right)+\left(A-B\right)$

Applying $\tan$ on both sides, we get,

$\begin{array}{ccccc}& \tan \left(2A\right)& =& \tan \left(\left(A+B\right)+\left(A-B\right)\right)& \\ & & =& \frac{\tan \left(A+B\right)+\tan \left(A-B\right)}{1-\tan \left(A+B\right)\tan \left(A-B\right)}& ;\left[\because \tan \left(x+y\right)=\frac{\tan \left(x\right)+\tan \left(y\right)}{1-\tan \left(x\right)\tan \left(y\right)}\right]\\ \Rightarrow & \tan \left(2A\right)& =& \frac{p+q}{1-pq}& ;\left[\because \tan \left(A+B\right)=p;\tan \left(A-B\right)=q\right]\end{array}$

Hence the correct option is option(C) i.e. $\frac{p+q}{1-pq}$