If sec x + tan x = k, cos x =

(b) $\frac{2 k}{k^{2}+1}$

We have:

$\sec x+\tan x=k$       ...(1)

$\Rightarrow \frac{1}{\sec x+\tan x}=\frac{1}{k}$

$\Rightarrow \frac{\sec ^{2} x-\tan ^{2} x}{\sec x+\tan x}=\frac{1}{k}$

$\Rightarrow \frac{(\sec x+\tan x)(\sec x-\tan x)}{(\sec x+\tan x)}=\frac{1}{k}$

$\therefore \sec x-\tan x=\frac{1}{k}$       ...(2)

Adding (1) and (2):

$2 \sec x=k+\frac{1}{k}$

$\Rightarrow 2 \sec x=\frac{k^{2}+1}{k}$

$\Rightarrow \sec x=\frac{k^{2}+1}{2 k}$

$\Rightarrow \frac{1}{\cos x}=\frac{k^{2}+1}{2 k}$

$\Rightarrow \cos x=\frac{2 k}{k^{2}+1}$