MCQ
જો $2{\tan ^2}\theta = {\sec ^2}\theta , $ તો $\theta $ નો વ્યાપક ઉકેલ મેળવો.
- A$n\pi + \frac{\pi }{4}$
- B$n\pi - \frac{\pi }{4}$
- ✓$n\pi \pm \frac{\pi }{4}$
- D$2n\pi \pm \frac{\pi }{4}$
$\Rightarrow 2{\tan ^2}\theta = {\tan ^2}\theta + 1$
$ \Rightarrow $ ${\tan ^2}\theta = 1 = {\tan ^2}\left( {\frac{\pi }{4}} \right)$
$\Rightarrow \theta = n\pi \pm \frac{\pi }{4}$.
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