Question

If tan θ + sec θ =e^x, then cos θ equals

(a) $\frac{e^x+e^{-x}}{2}$
(b) $\frac{2}{e^x+e^{-x}}$
(c) $\frac{e^x-e^{-x}}{2}$
(d) $\frac{e^x-e^{-x}}{e^x+e^{-x}}$

Answer 1

(b) $\frac{2}{e^x+e^{-x}}$

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