# An odd trig identity, I WANT PROOF!

When I was checking my work, Wolframalpha took my trig work a step further with an identity that no one in my Calculus II class has ever seen, including my teacher.

csc(2x) - cot(2x) = tan(x)

I tried to prove the identity myself and I looked online, but no luck. Please, could someone prove this?

We have

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

Can you prove this?? Can you take it from there??

Just write this as $\displaystyle\dfrac{ 1}{\sin\left(2\cdot x\right)} -\dfrac{\cos\left(2\cdot x\right)}{ \sin\left(2\cdot x\right)}=\dfrac{ \sin\left(x\right)}{\cos\left(x\right)}$ while remembering that $\sin^2\left(x\right)+\cos^2\left(x\right)=1$. Then apply the standard trig identities.

Oh, and, is this really calculus?

Just write this as $\displaystyle\dfrac{ 1}{\sin\left(2\cdot x\right)} -\dfrac{\cos\left(2\cdot x\right)}{ \sin\left(2\cdot x\right)}=\dfrac{ \sin\left(x\right)}{\cos\left(x\right)}$ while remembering that $\sin^2\left(x\right)+\cos^2\left(x\right)=1$. Then apply the standard trig identities.

Oh, and, is this really calculus?

This isn't really Calculus; however, you do have to know your trig identities with various questions.