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

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

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

We have:

$tan\theta +sec\theta =e^x$

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

$\Rightarrow \frac{1}{sec\theta +tan\theta }=\frac{1}{e^x}$

$\Rightarrow \frac{se{c}^2\theta -ta{n}^2\theta }{sec\theta +tan\theta }=\frac{1}{e^x}$

$\Rightarrow \frac{(sec\theta +tan\theta )(sec\theta -tan\theta )}{(sec\theta +tan\theta )}=\frac{1}{e^x}$

$\therefore sec\theta -tan\theta =\frac{1}{e^x}.....\left(2\right)$

Adding (1)and (2):

$2sec\theta =e^x+\frac{1}{e^x}$

$\Rightarrow 2sec\theta =\frac{(e^x{)}^2+1}{e^x}$

$\Rightarrow sec\theta =\frac{e^{2x}+1}{2e^x}$

$\Rightarrow sec\theta =\frac{1}{2}\times \frac{e^{2x}+1}{e^x}$

$\Rightarrow sec\theta =\frac{1}{2}\times (e^x+e^{-x})$

$\Rightarrow \frac{1}{cos\theta }=\frac{e^x+e^{-x}}{2}$

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