If secA + tanA = a, then the value of cosA is.

A. \(\frac{{{a^2}}}{{2a}}\)

B. \(\frac{{2a}}{{{a^2}\; + \;1}}\)

C. \(\frac{{{a^2} - 1}}{{2a}}\)

D. \(\frac{a}{{{a^2} - 1}}\)

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Given: secA + tanA = a ----(1)

⇒ we know that sec²A – tan² A = 1

⇒ (secA + tanA) (secA – tanA) = 1

⇒ a(secA – tanA) = 1

⇒ (secA – tanA) = 1/a ----(2)

⇒ Adding equation 1 and 2 we get

⇒ 2secA = a + (1/a)

⇒ 2secA = (a² + 1) /a

∴ cosA = 2a/ (a² + 1)