If + =e^x, then =

MCQ

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

A
$\frac{\text{e}^\text{x}+\text{e}^{-\text{x}}}{2}$
B
$\frac{2}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$
C
$\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{2}$
D
$\frac{\text{e}^\text{x}-\text{e}^{-\text{x}}}{\text{e}^\text{x}+\text{e}^{-\text{x}}}$

✓

Answer

Solution:

We have:

$\tan\theta + \sec\theta=\text{e}^\text{x}$

$\sec\theta + \tan\theta = \text{e}^\text{x}\cdots(1)$

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

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

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

$\therefore \sec\theta - \tan\theta=\frac{1}{\text{e}^\text{x}}\cdots(2)$

Adding (1) and (2):

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

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

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

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

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

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

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

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