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3.Trigonometrical Ratios, Functions and Identities
easy
If $\sec \theta + \tan \theta = p,$ then $\tan \theta $ is equal to
A
$\frac{{2p}}{{{p^2} - 1}}$
B
$\frac{{{p^2} - 1}}{{2p}}$
C
$\frac{{{p^2} + 1}}{{2p}}$
D
$\frac{{2p}}{{{p^2} + 1}}$
Solution
(b) $\sec \theta + \tan \theta = p\,\, $……..$(i)$
$\Rightarrow \,\sec \,\theta – \tan \theta = \frac{1}{p}$……….$(ii)$
Subtracting second from first, we get
$2\tan \theta = p – \frac{1}{p}$
$ \Rightarrow \,\tan \theta = \frac{{{p^2} – 1}}{{2p}}$.
Standard 11
Mathematics
