Question

Tan x=-1. please explain

Answer 1

If $-\dfrac\pi2$ (which is the domain of $\tan x$), then $\tan x=-1$ for $x=\tan^{-1}(-1)=-\tan^{-1}1=-\dfrac\pi4$.
But also, recall that $\tan x$ has period $\pi$, which means $\tan(x+n\pi)=\tan x$ for all integers $n$. This means the general solution to $\tan x=-1$ is $-\dfrac\pi4+n\pi$ for $n\in\mathbb Z$.

This doesn't match any of the given choices, but we can simply write

$x=-\dfrac\pi4+n\pi=-\dfrac\pi4+\pi+(n-1)\pi=\dfrac{3\pi}4+(n-1)\pi$

and we can just replace $n-1$ with $n$, since both can be arbitrary integers, which means (D) is the correct answer.