3.Trigonometrical Ratios, Functions and Identities
easy

यदि $\sec \theta  + \tan \theta  = p,$ तब $\tan \theta $ बराबर है

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

$\sec \theta  + \tan \theta  = p$…..$(i)$

$\sec \theta  – \tan \theta  = \frac{1}{p}$…..$(ii)$

$(ii)$ को $(i)$ से घटाने पर,

$2\tan \theta  = p – \frac{1}{p}$

$ \Rightarrow \,\tan \theta  = \frac{{{p^2} – 1}}{{2p}}$.

Standard 11
Mathematics

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