Question

Evaluate $\tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right)$

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

Answer 1

The correct option is A

$2\tan \left(2\theta \right)$

Explanation for the correct option:

Evaluating the given expression:

Given expression is $\tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right)$.

We know that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Applying this identity, we get

$$\begin{gathered} \tan \left(\frac{\pi }{4}+\theta \right)-\tan \left(\frac{\pi }{4}-\theta \right) \\ =\frac{\tan \left(\frac{\pi }{4}\right)+\tan \left(\theta \right)}{1-\tan \left({\displaystyle \frac{\pi }{4}}\right)\tan \left(\theta \right)}-\frac{\tan \left(\frac{\pi }{4}\right)-\tan \left(\theta \right)}{1+\tan \left({\displaystyle \frac{\pi }{4}}\right)\tan \left(\theta \right)} \\ =\left[\frac{1+\tan \left(\theta \right)}{1-\tan \left(\theta \right)}\right]-\left[\frac{1-\tan \left(\theta \right)}{1+\tan \left(\theta \right)}\right]\mathbf{}\left[\because tan\left(\frac{\pi }{4}\right)=1\right] \\ =\left[\frac{{(1+\tan \left(\theta \right))}^2-{(1-\tan \left(\theta \right))}^2}{1-{\tan }^2\left(\theta \right)}\right] \\ =\frac{4\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}\mathbf{}\left[{\because (a+b)}^2–{(a–b)}^2=4ab\right] \\ =\frac{2\left(2\tan \right(\theta \left)\right)}{1-{\tan }^2\left(\theta \right)} \\ =2\tan \left(2\theta \right)\left[\because \frac{2tan\left(\theta \right)}{1-ta{n}^2\left(\theta \right)}=tan\left(2\theta \right)\right] \end{gathered}$$

Hence, the solution is $2\tan \left(2\theta \right)$.

Therefore, option(A) is the correct answer.