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

Answer: (B)

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

$\tan \theta +\sec \theta =e^x$
$\Rightarrow \frac{1+\sin \theta }{\cos \theta }=e^x\Rightarrow \frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}=e^x$
(U\sin g componendo & Dividendo rule)
$\Rightarrow \tan \frac{\theta }{2}=\frac{e^x-1}{e^x+1}$
as $\cos \theta =\frac{1-{\tan }^2\frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}$
$\Rightarrow \cos \theta =\frac{1-{\left(\frac{e^x-1}{e^x+1}\right)}^2}{1+{\left(\frac{e^x-1}{e^x+1}\right)}^2}$
$\Rightarrow \cos \theta =\frac{2e^x}{e^{2x}+1}$
$\Rightarrow \cos \theta =\frac{2}{e^x+e^{-x}}$.
OPTION B