If sec x + tan x = k, cos x =(a) (k² + 1)/2k(b) 2k/(k² + 1)(c) k/(k² + 1)(d) k/(k² − 1)

Question

\[ \text{If } \sec x+\tan x=k, \]

\[ \text{then } \cos x= \]

(a) \(\dfrac{k^2+1}{2k}\)
(b) \(\dfrac{2k}{k^2+1}\)
(c) \(\dfrac{k}{k^2+1}\)
(d) \(\dfrac{k}{k^2-1}\)

Using identity

\[ (\sec x+\tan x)(\sec x-\tan x)=1 \]

\[ \sec x-\tan x=\frac1k \]

Adding,

\[ 2\sec x = k+\frac1k \]

\[ \sec x = \frac{k^2+1}{2k} \]

\[ \cos x = \frac{2k}{k^2+1} \]

\[ \boxed{\frac{2k}{k^2+1}} \]

Correct Option: (b)

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