Question

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

The correct option is B $\frac{2}{(e^x+e^{-x})}$

$\tan \theta +\sec \theta =e^x$.....(1).

As ${\sec }^2\theta -{\tan }^2\theta =1$ we have

$\sec \theta -\tan \theta =e^{-x}$......(2).

Now adding (1) and (2) we get,

$\sec \theta =\frac{e^x+e^{-x}}{2}$

or, $\cos \theta =\frac{2}{e^x+e^{-x}}$