Question

A general solution of ${\tan }^2\theta +\cos 2\theta =1$ is $\left(nϵz\right)$

• A. $n\pi -\frac{\pi }{4}$
• B. $2n\pi +\frac{\pi }{4}$
• C. $n\pi +\frac{\pi }{4}$
• D. $n\pi$

Answer 1

Answer: (B), (D)

The correct options are
B $n\pi$
D $2n\pi +\frac{\pi }{4}$

${\tan }^2\theta +\cos 2\theta =1$
${\tan }^2\theta +\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=1$
${\tan }^2\theta +{\tan }^4\theta +1-{\tan }^2\theta =1+{\tan }^2\theta$
${\tan }^4\theta -{\tan }^2\theta =0$
${\tan }^2\theta ({\tan }^2\theta -1)=0$
$\tan \theta =0$ or $\tan \theta =\pm 1$
So,
$\theta =n\pi$ or $2n\pi +\frac{\pi }{4}$