Proving $1+\cot^2(-\theta)=\csc^2(\theta)$ I'm stuck on this one proof that I just can't get for some reason. It seems really simple too, and I've tried just about everything I can think of, but I just keep going in circles.

$$1+\cot^2(-\theta)=\csc^2(\theta)$$

I know that $1 + \cot^2(-\theta) = 1 - \cot^2(\theta)$; that is, that the function is odd, so $f(-x) = -f(x)$.
From there, I've tried a bunch of stuff - too much to list, but a few of them are (working with LHS):
1.) Replacing $\cot^2\theta$ by $\frac{1}{\tan^2(\theta)}$ 
2.) Replace $1$ by $\csc^2\theta - \cot^2\theta$ 
3.) Multiplying LHS by $\frac{\sin(\theta)}{\sin(\theta)}$ (Don't even remember why I tried this, I was just frustrated)
Could someone please point me in the right direction? This is driving me nuts.
 A: $1+\cot^2(-\theta)=1+(-\cot\theta)^2=1+\cot^2\theta=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{1}{\sin^2\theta}=\csc^2\theta$
A: You were on the right path:
$$\begin{align}1+\cot^2(\theta)
&=\frac{\sin^\color{green}2(\theta)}{\sin^\color{green}2(\theta)}(1+\cot^2(\theta))\\
&=\frac{\sin^2(\theta)+\sin^2(\theta)\cot^2(\theta)}{\sin^2(\theta)}\\
&=\frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)}\\
&=\frac1{\sin^2(\theta)}\\
&=\csc^2(\theta).\end{align}$$
A: Initially note that 
$$\cot^2(-\theta)=(\cot(-\theta))^2=(-\cot\theta)^2=(\cot\theta)^2=\cot^2\theta$$
Then, proof the identity $1+\cot^2(-\theta)= \csc^2\theta$ is same that proof
$$1+\cot^2\theta=\csc^2\theta$$
Hence, divide the identity $\sin^2\theta+\cos^2\theta=1$ by $\sin^2\theta$.
$$\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta} \Rightarrow 1+\cot^2\theta=\csc^2\theta.$$
A: One way to look at these is to convert all expressions to ratios of the variables $o,a,h$
$$1+cot^2(-θ)=csc^2(θ)$$
$$1+(\frac{a}{o})^2=(\frac{h}{o})^2$$
$$(\frac{o}{o})^2+(\frac{a}{o})^2=(\frac{h}{o})^2$$
$$\frac{o^2+a^2}{o^2}=\frac{h^2}{o^2}$$
$$o^2+a^2=h^2$$
Which checks out due to the Pythagorean Theorem. I skipped over the negative angle identity of cotan since it is squared.
A: Your attempt to solve the problem by replacing $1$ by $\csc^2\theta - \cot^2\theta$ was a good idea.  Since $\cot\theta$ is an odd function, $\cot(-\theta) = -\cot\theta$.  Thus,
\begin{align*}
1 + \cot^2(-\theta) & = 1 + (-\cot\theta)^2\\
                   & = 1 + \cot^2\theta\\
                   & = \csc^2\theta - \cot^2\theta + \cot^2\theta\\
                   & = \csc^2\theta
\end{align*}
Where you made your mistake was in concluding that the fact that cotangent is odd implied $1 + \cot^2(-\theta) = 1 - \cot^2\theta$ rather than $1 + \cot^2(-\theta) = 1 + (-\cot\theta)^2 = 1 + \cot^2\theta$.
