# Verify the trig identity $\frac{\cot^2\theta-1}{\csc^2\theta}=\csc \theta -1$

I've stumbled across a brain-teaser. After using some identities, I get the left hand side equal to $$\cos^2\theta - \sin^2\theta$$. I'm not aware of any other identities that could get me to the right hand side. I'm actually leaning towards there being a typo in the problem.

• That's definitely not an identity: only the LHS is bounded. Maybe you're meant to find which $\theta$ satisfy it as an equation? – J.G. Nov 25 '19 at 22:15
• Yeah, something seems off. Especially if you got the LHS to equal cos^2 - sin^2, since that's equal to cos(2\theta) which is definitely not the same as csc(theta)-1. – ConMan Nov 25 '19 at 22:18
• Plug in x=pi/2 for example，it's not true – Gabe Nov 25 '19 at 22:18
• That said, based on a quick graph on Wolfram Alpha, I don't think the two sides ever equal each other, either. – ConMan Nov 25 '19 at 22:18

## 2 Answers

Yes indeed we have

$$\frac{\cot^2\theta-1}{csc^2\theta}=\frac{\frac{\cos^2\theta}{\sin^2\theta}-1}{\frac{1}{\sin^2\theta}}=\cos^2 \theta-\sin^2\theta =\cos (2\theta)\neq csc \theta -1$$

If $$\cos\2theta=\csc\theta-1$$

As $$\sin\theta\ne0,$$

Multiply both sides by $$2\sin\theta$$ and use http://mathworld.wolfram.com/WernerFormulas.html

$$\sin3\theta-\sin\theta=2-2\sin\theta$$

$$2=\sin3\theta+\sin\theta$$

We need $$\sin3\theta=\sin\theta=1$$

but $$\sin3\theta=3\sin\theta-4\sin\theta=3-4=-1$$

Alternatively using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$2=\sin3\theta+\sin\theta=2\sin2\theta\cos\theta$$

We need $$\sin2\theta=\cos\theta=\pm1$$

But $$\cos\theta=\pm1\implies\sin2\theta=0$$