If tan θ + sec θ =e

Solution:

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

We have:

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

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

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

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

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

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

Adding $(1)$ and $(2):$

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

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

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

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

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

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

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