Question

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

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

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

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

(d) $\frac{k}{k^2-1}$

Answer 1

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

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \sec x+\tan x=k\left(1\right) \\ \Rightarrow \frac{1}{\sec x+\tan x}=\frac{1}{k} \\ \Rightarrow \frac{{\sec }^{2}x-{\tan }^{2}x}{\sec x+\tan x}=\frac{1}{k} \\ \Rightarrow \frac{\left(\sec x+\tan x\right)\left(\sec x-\tan x\right)}{\left(\sec x+\tan x\right)}=\frac{1}{k} \\ \therefore \mathrm{s}\mathrm{e}\mathrm{c}x-\tan x=\frac{1}{k}\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right): \\ 2\sec x=k+\frac{1}{k} \\ \Rightarrow 2\sec x=\frac{k^2+1}{k} \\ \Rightarrow \sec x=\frac{k^2+1}{2k} \\ \Rightarrow \frac{1}{\cos x}=\frac{k^2+1}{2k} \\ \Rightarrow \cos x=\frac{2k}{k^2+1} \end{gathered}$$