Q. General solution of the equation $tan^{2}\theta +sec2\theta =1$ is

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Solution:

Using, $sec2\theta =\frac{1}{c o s 2 \theta }=\frac{1 + t a n^{2} \theta }{1 - t a n^{2} \theta }$
$\Rightarrow $ We can write the given equation as $tan^{2}\theta +\frac{1 + t a n^{2} \theta }{1 - t a n^{2} \theta }=1$
$\Rightarrow $ $tan^{2}\theta \left(\right.1-tan^{2}\theta \left.\right)+1+tan^{2}\theta =1-tan^{2}\theta $
$\Rightarrow $ $3tan^{2}\theta -tan^{4}\theta =0\Rightarrow tan^{2}\theta \left(\right.3-tan^{2}\theta \left.\right)=0$
$\Rightarrow \tan \theta=0 \Rightarrow \theta=m \pi, m \in I$
$\Rightarrow \text{tan\theta } = 0 \Rightarrow \text{\theta } = m \text{\pi ,} \, m \in I$
or $tan^{2} \theta = tan^{2} \frac{\pi }{3}$ $\Rightarrow \theta =n\pi \pm\frac{\pi }{3}$ , $n \in I$