MCQ
જો $\tan \theta = t,$ તો $\tan 2\theta + \sec 2\theta = $
- ✓$\frac{{1 + t}}{{1 - t}}$
- B$\frac{{1 - t}}{{1 + t}}$
- C$\frac{{2t}}{{1 - t}}$
- D$\frac{{2t}}{{1 + t}}$
$\tan 2\theta + \sec 2\theta = \frac{{2t}}{{1 - {t^2}}} + \frac{{1 + {t^2}}}{{1 - {t^2}}} $
$= \frac{{{{(1 + t)}^2}}}{{(1 - t)(1 + t)}} = \frac{{1 + t}}{{1 - t}}$.
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