Question

If $\tan \left(A\right)=2\tan \left(B\right)+cot\left(B\right)$, then $2\tan \left(A-B\right)$ is equal to

• A. $cot\left(B\right)$$\tan \left(B\right)$
• B. $2\tan \left(B\right)$
• C. $cot\left(B\right)$
• D. $2cot\left(B\right)$

Answer 1

The correct option is C

$cot\left(B\right)$

Explanation for the correct option:

Step 1: Given that,

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

$\begin{array}{rcl}\tan \left(A\right)& =& 2\tan \left(B\right)+\frac{1}{\tan \left(B\right)}\\ \tan \left(A\right)& =& \frac{2{\tan }^2\left(B\right)+1}{\tan \left(B\right)}\\ \tan \left(A\right)\tan \left(B\right)& =& 2{\tan }^2\left(B\right)+1...\left(i\right)\end{array}$

Step 2: Evaluate$\tan \left(A\right)-\tan \left(B\right)$ the given equation.

$\begin{array}{c}\therefore \tan \left(A\right)-\tan \left(B\right)=2\tan \left(B\right)+\cot \left(B\right)-\tan \left(B\right)\\ =\tan \left(B\right)+\cot \left(B\right)\\ =\tan \left(B\right)+\frac{1}{\tan \left(B\right)}\\ =\frac{{\tan }^2\left(B\right)+1}{\tan \left(B\right)}...\left(ii\right)\end{array}$

$\begin{array}{c}\text{2 tan}\left(\text{A–B}\right)\text{=2}\left[\frac{\left(\text{tan}\left(\text{A}\right)\text{– tan}\left(\text{B}\right)\right)}{\left(\text{1 + tan}\left(\text{A}\right)\text{tan}\left(\text{B}\right)\right)}\right]\\ \text{= 2}\frac{\left[\frac{\left({\text{tan}}^{\text{2}}\left(\text{B}\right)\text{+ 1}\right)}{\text{tan}\left(\text{B}\right)}\right]}{\left({\text{1 + 2 tan}}^{\text{2}}\left(\text{B}\right)\text{+ 1}\right)}\left\{\text{from}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\right\}\\ \text{=}\frac{\left[\text{2}\left({\text{1 + tan}}^{\text{2}}\left(\text{B}\right)\right)\right]}{\left[\text{2}\left({\text{1 + tan}}^{\text{2}}\left(\text{B}\right)\right)\text{tan}\left(\text{B}\right)\right]}\\ \text{=}\frac{\text{1}}{\text{tan}\left(\text{B}\right)}\\ \text{= cot}\left(\text{B}\right)\end{array}$

Hence, the correct option is (C).