Question

Solve $4\cot 2\theta ={\cot }^2\theta -{\tan }^2\theta$

• A. $\theta =n\pi \pm \frac{\pi }{4}$, $nϵZ$
• B. $\theta =n\pi \pm \frac{\pi }{2}$, $nϵZ$
• C. $\theta =n\pi \pm \frac{\pi }{3}$, $nϵZ$
• D. $\theta =n\pi \pm \frac{\pi }{8}$, $nϵZ$

Answer 1

The correct option is A $\theta =n\pi \pm \frac{\pi }{4}$, $nϵZ$

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

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

Substitute

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

$(1-{\tan }^2\theta )[2\tan \theta -(1+{\tan }^2\theta )]=0$

$(1-{\tan }^2\theta )({\tan }^2\theta -2\tan \theta +1)=0$

$(1-{\tan }^2\theta ){(\tan \theta -1)}^2=0$

$\tan \theta =\pm 1$

$\Rightarrow$ $\theta =n\pi \pm \frac{\pi }{4}$, $n\in Z$