Question

General value of $\theta$ satisfying equation ${\tan }^2\theta +\sec 2\theta =1$ is

• A. $n\pi$
• B. $n\pi +\frac{\pi }{3}$
• C. $n\pi -\frac{\pi }{3}$
• D. All of these

Answer 1

The correct option is D All of these

We know that ,

$text{sec}^2\theta ={\tan }^2\theta +1$

$\text{sec}2\theta +{\text{sec}}^2\theta -1=1$
$\frac{1}{2{\cos }^2\theta -1}=2-{\text{sec}}^2\theta$

$1=(2{\cos }^2\theta -1)(2-{\text{sec}}^2\theta )$

$4{\cos }^4\theta -5{\cos }^2\theta +1=0$

$\cos \theta =-1,-(\frac{1}{2}),(\frac{1}{2}),1$

general solutions are ;

$\theta =n\pi ,n\pi -\frac{\pi }{3},n\pi +\frac{n\pi }{3}$