Question

The general solution of ${\tan }^2\theta +{\cot }^2\theta =2$ is

• A. $0=n\pi \pm \frac{\pi }{4};n\in Z$
• B. $0=n\pi \pm \frac{\pi }{3};n\in 1$
• C. $\theta =n\pi \pm \frac{\pi }{6};n\in I$
• D. $0=n\pi \pm \frac{\pi }{5};n\in 1$

Answer 1

The correct option is A $0=n\pi \pm \frac{\pi }{4};n\in Z$

Given ${\tan }^2\theta +{\cot }^2\theta =2$

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

$⟹(\tan \theta {)}^2+(\cot \theta {)}^2-2\left(\tan \theta \right)\left(\cot \theta \right)=0$

$⟹(\tan \theta -\cot \theta {)}^2=0$

$⟹\tan \theta -\frac{1}{\tan \theta }=0$

$⟹{\tan }^2\theta =1$

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