Question

If sec x + tan x = k, cos x =

Options

• \[\frac{k^2 + 1}{2k}\]

   

• \[\frac{2k}{k^2 + 1}\]

   

• \[\frac{k}{k^2 + 1}\]

   

• \[\frac{k}{k^2 - 1}\]

   

Answer 1

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

We have:

\[\sec x + \tan x = k \left( 1 \right)\]

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

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

\[ \Rightarrow \frac{\left( \sec x + \tan x \right)\left( \sec x - \tan x \right)}{\left( \sec x + \tan x \right)} = \frac{1}{k}\]

\[ \therefore \sec x-\tan x = \frac{1}{k} \left( 2 \right)\]

Adding ( 1 ) and ( 2 ): 

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

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

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

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

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