Question

State true or false $\tan (\frac{\pi }{4}+\theta )-\tan (\frac{\pi }{4}-\theta )=2\tan \theta$

• A. True
• B. False

Answer 1

The correct option is B False

Using $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\Rightarrow \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A+\tan B}$

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

$\Rightarrow A=\frac{\pi }{4},B=\theta$

$\Rightarrow \frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta }-\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\tan \theta }$

$\because \tan \frac{\pi }{4}=1$

$\Rightarrow \frac{1+\tan \theta }{1-\tan \theta }-\frac{1-\tan \theta }{1+\tan \theta }$

$\Rightarrow \frac{(1+\tan \theta {)}^2-(1-\tan \theta {)}^2}{(1-\tan \theta )(1+\tan \theta )}$

$\therefore a^2-b^2=(a+b)(a-b)$

$\Rightarrow \frac{1+{\tan }^2\theta +2\tan \theta -1-{\tan }^2\theta +2\tan \theta }{1-{\tan }^2\theta }$

$\Rightarrow \frac{4\tan \theta }{1-{\tan }^2\theta }$

$\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$

$=2\tan 2\theta$.