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Question
If sec x + tan x = k, cos x =
Options
- \[\frac{k^2 + 1}{2k}\]
- \[\frac{2k}{k^2 + 1}\]
- \[\frac{k}{k^2 + 1}\]
- \[\frac{k}{k^2 - 1}\]
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Solution
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 ):
\[2\sec x = k + \frac{1}{k}\]
\[ \Rightarrow 2\sec x = \frac{k^2 + 1}{k}\]
\[ \Rightarrow \sec x = \frac{k^2 + 1}{2k}\]
\[ \Rightarrow \frac{1}{\cos x} = \frac{k^2 + 1}{2k}\]
\[ \Rightarrow \cos x = \frac{2k}{k^2 + 1}\]
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