Answer
$e-e^{-1}$
Work Step by Step
As we are given that $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta$
Now, consider $\tan \theta=t$ and $dt= \sec^2 \theta d\theta$
Thus, $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta=\int^{1}_{-1} \cosh t dt=[\sin h t]^{1}_{-1}$
and, $=2 \sinh (1)$
As we know $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$
Then $= 2[\dfrac{e^{1} -e^{-1}}{2}]$
Hence, $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta=e-e^{-1}$
