Question

The complete solution set of $4\cot 2\theta ={\cot }^2\theta -{\tan }^2\theta$ is

• A. $\theta =n\pi +\frac{\pi }{4},n\in Z$
• B. $\theta =n\pi -\frac{\pi }{4},n\in Z$
• C. $\theta =n\pi \pm \frac{\pi }{8},n\in Z$
• D. $\theta =n\pi \pm \frac{\pi }{4},n\in Z$

Answer 1

The correct option is D $\theta =n\pi \pm \frac{\pi }{4},n\in Z$

$\frac{4}{\tan 2\theta }=\frac{1}{{\tan }^2\theta }-{\tan }^2\theta \Rightarrow \frac{4(1-{\tan }^2\theta )}{2\tan \theta }=\frac{1-{\tan }^4\theta }{{\tan }^2\theta }\Rightarrow \frac{(1-{\tan }^2\theta )[2\tan \theta -(1+{\tan }^2\theta \left)\right]}{{\tan }^2\theta }=0[\because \tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }]\Rightarrow \frac{({\tan }^2\theta -1)(\tan \theta -1{)}^2}{{\tan }^2\theta }=0\Rightarrow {\tan }^2\theta =1,\tan \theta =1\therefore \theta =n\pi \pm \frac{\pi }{4},n\in Z$