Answer
$$\lim_{\theta\to\pi/4}\frac{\tan\theta-1}{\theta-\frac{\pi}{4}}=2$$
Work Step by Step
$$A=\lim_{\theta\to\pi/4}\frac{\tan\theta-1}{\theta-\frac{\pi}{4}}=\lim_{\theta\to\pi/4}\frac{\tan\theta-\tan\frac{\pi}{4}}{\theta-\frac{\pi}{4}}$$
Let a function $f(x)=\tan x$. According to the definition of derivative, we have
$$f'(x)=\lim_{z\to x}\frac{\tan z-\tan x}{z-x}=(\tan x)'=\sec^2 x$$
So we take $z=\theta$ and $x=\pi/4$ for the limit in this exercise, we have
$$A=\lim_{\theta\to\pi/4}\frac{\tan\theta-\tan\frac{\pi}{4}}{\theta-\frac{\pi}{4}}=\sec^2\Big(\frac{\pi}{4}\Big)=\frac{1}{\cos^2\frac{\pi}{4}}$$
$$A=\frac{1}{\frac{(\sqrt2}{2})^2}=\frac{1}{\frac{1}{2}}=2$$
