Question

If sec θ + tan θ = k, cos θ =

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

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