MCQ
If $\tan \theta = t,$ then $\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}}} $
$= \frac{{{{(1 + t)}^2}}}{{(1 - t)(1 + t)}} = \frac{{1 + t}}{{1 - t}}$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$ =\frac{1}{2024}, $ then $\alpha$ is equal to-