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.
Best Answer
$1+\cot^2(-\theta)=1+(-\cot\theta)^2=1+\cot^2\theta=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{1}{\sin^2\theta}=\csc^2\theta$