Solve for $x$: $\csc^{100}x+\tan^{100}x=1$ 
Solve for $x$: $$\mathrm{csc}^{100}x+\tan^{100}x=1$$

I have tried it so many times but couldn't draw any conclusion. Please help.
 A: According to the Mark Bennet comment: $$\csc x\ge 1 ,\,\,\,\,\,\forall x\in\mathbb{R}.$$ Therefore for real solutions of this trig equation, we need $\tan x=0$ and $\csc x=1.$ If such a solution exist, then$$\sin x=0=1$$ which is a contradiction. Therefore there are no such real $x.$
A: We get,
$$-1\leq \sin(x) \leq 1 \implies 0\leq \sin^2(x) \leq 1 \implies 1\leq \csc^2(x) < \infty $$$$\implies 1\leq \csc^{100}(x) < \infty , \hspace{.3cm}\forall x \in \mathbb{R} $$
and
$$-\infty < \tan(y)<\infty \implies 0\leq \tan^2(y)<\infty \implies 0\leq \tan^{100}(y)<\infty ,  \hspace{0.3cm}\forall y \in \mathbb{R}.$$

So the only possibility for, $\csc^{100}(x)+\tan^{100}(x)=1$ is the existence of an $x \in \mathbb{R}$ such that $\csc^{100}(x)=1$ and $\tan^{100}(x)=0$ , $ i.e., \sin(x)=\pm 1$ ($\implies x$ is an odd integer multiple of $\frac {\pi}{2}$) and $\tan(x)=0$ ($\implies x$ is an even integer multiple of $ {\pi}$), which is not possible.
A: Note that the given equation suggests $$\mathrm{cosec} x\le 1$$ whihc suggests the only possible solution as $\mathrm{cosec} x=1\Rightarrow x=\frac{\pi}{2} $ which is impossible because that makes $\tan x =\infty$.
