Question

If $\tan \, \theta + \sec \, \theta = e^x$ , then $\cos \, \theta$ equal is

• 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

Answer: (B)

Here, $\sec \, \theta + \tan \, \theta = e^x$ ....(i)
Also $\sec^2 \, \theta - \tan^2 \, \theta = 1$ ....(ii)
Dividing (it) by (i), we get
$\sec \, \theta - \tan \, \theta = e^{-x}$ ...(iii)
Add (i) and (iii) to obtain $\sec \, \theta = \frac{e^x + e^{-x}}{2}$