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
    $\frac{{2t}}{{1 + t}}$

Answer

Correct option: A.
$\frac{{1 + t}}{{1 - t}}$
a
(a) $\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }},\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

$\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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