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3.Trigonometrical Ratios, Functions and Identities
easy
If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $
A
$\frac{{1 + t}}{{1 - t}}$
B
$\frac{{1 - t}}{{1 + t}}$
C
$\frac{{2t}}{{1 - t}}$
D
$\frac{{2t}}{{1 + t}}$
Solution
(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}}} $
$= \frac{{{{(1 + t)}^2}}}{{(1 – t)(1 + t)}} = \frac{{1 + t}}{{1 – t}}$.
Standard 11
Mathematics
