Question

The general solution of the equation $\tan (\frac{\pi }{4}+\theta )+\tan (\frac{\pi }{4}-\theta )=4$ is

• A. $n\pi \pm \frac{\pi }{4}$
• B. $2n\pi \pm \frac{\pi }{6}$
• C. $n\pi \pi \frac{\pi }{4}$
• D. $n\pi +(-1{)}^n\frac{\pi }{4}$

Answer 1

The correct option is A $n\pi \pm \frac{\pi }{4}$

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

$\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }=4$

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

$\frac{2(1+{\tan }^2\theta )}{1-{\tan }^2\theta }=4$

$1+{\tan }^2\theta =2-2{\tan }^2\theta$

$1+{\tan }^2\theta =2-2{\tan }^2\theta$

$3{\tan }^2\theta =1$

$\tan \theta =\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}$

$\tan \theta =\tan \frac{\pi }{3},\tan (-\frac{\pi }{3})$

$\theta =n\pi \pm \frac{\pi }{3}$