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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.

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  • $\begingroup$ just to be sure, are points 1.) and 2.) the correct replacements you want to do? $cot^{\theta}$ and $csc^{\theta}$? $\endgroup$ – iadvd Jun 24 '15 at 3:08
  • $\begingroup$ Using the Pythagorean identity, divide both sides by $\sin^2{\theta}$. $\endgroup$ – Brian Diaz Jun 24 '15 at 3:10
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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 $1 + \cot^2(-\theta) = 1 + (-\cot\theta)^2 = 1 + \cot^2\theta$.

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  • $\begingroup$ Yes, I noticed my mistake as soon as I saw everyone's answers! Hit me like a ton of bricks, haha. Thanks for the help, everyone! $\endgroup$ – enharmonics Jun 24 '15 at 13:38
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$1+\cot^2(-\theta)=1+(-\cot\theta)^2=1+\cot^2\theta=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}=\dfrac{1}{\sin^2\theta}=\csc^2\theta$

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You were on the right path:

$$\begin{align}1+\cot^2(\theta) &=\frac{\sin^\color{green}2(\theta)}{\sin^\color{green}2(\theta)}(1+\cot^2(\theta))\\ &=\frac{\sin^2(\theta)+\sin^2(\theta)\cot^2(\theta)}{\sin^2(\theta)}\\ &=\frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^2(\theta)}\\ &=\frac1{\sin^2(\theta)}\\ &=\csc^2(\theta).\end{align}$$

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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.

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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 $\sin^2\theta$.

$$\frac{\sin^2\theta}{\sin^2\theta}+\frac{\cos^2\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta} \Rightarrow 1+\cot^2\theta=\csc^2\theta.$$

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    $\begingroup$ The commands for the trigonometric functions, used in math mode, are \sin for sine, \cos for cosine, \tan for tangent, \csc for cosecant, \sec for secant, and \cot for cotangent. $\endgroup$ – N. F. Taussig Jun 24 '15 at 9:43

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