I thinking about how to write inverse trig functions as each other, eg writing arccsc as some form of arctan, etc. Can anyone tell me if what I did was right? If not why? Here is what I did:
Start with
$$\csc^{-1}(u)=y$$
so
$$\csc(y)=u$$
Draw a right-angled triangle from the acute reference angle $y$, that has hypotenuse side $u$ and opposite side $1$. So it follows that the adjacent side is $\sqrt{u^2-1}$.
From this we can see that:
$$\tan(y)=\frac{1}{\sqrt{u^2-1}}$$
$$y=\tan^{-1}\left(\frac{1}{\sqrt{u^2-1}}\right)$$
And so we have the identity:
$$\csc^{-1}(u)=\tan^{-1}\left(\frac{1}{\sqrt{u^2-1}}\right)$$
Is this identity I derived correct?
As Jjagmath pointed out, the identity you derived is correct only for $\,u>1\,,\,$ whereas you need to change the sign for $\,u<-1\,.$
The function $\,y=\csc(x)\,$ is invertible on $\,\left[-\dfrac{\pi}2,0\right)\bigcup\left(0,\dfrac{\pi}2\right]\,$ and its inverse function is
$x=\csc^{-1}(y):\!\big(\!-\!\infty,-1\big]\bigcup\big[1,+\infty\big)\!\to\!\left[-\dfrac{\pi}2,0\right)\bigcup\left(0,\dfrac{\pi}2\right].$
On the other hand, the function $\,y=\tan(x)\,$ is invertible on the interval $\,\left(-\dfrac{\pi}2,\dfrac{\pi}2\right)\,$ and its inverse function is
$x=\tan^{-1}(y):\big(\!-\infty,+\infty\big)\to\left(-\dfrac{\pi}2,\dfrac{\pi}2\right)\,.$
For any $\,y\in\big(\!-\infty,-1\big)\,,\,$ let $\,x=\csc^{-1}(y)\in\left(-\dfrac{\pi}2,0\right)\,.$
It results that $\;y=\csc(x)\,,\,$ $\sin x<0\,,\,$ $\cos x>0\,$ and
$\begin{align}\tan(x)&=\dfrac{\sin x}{\cos x}=\dfrac{\sin x}{\sqrt{1-\sin^2(x)}}=\dfrac{\sin x}{\left|\sin x\right|\sqrt{\dfrac1{\sin^2(x)}-1}}=\\&=-\dfrac1{\sqrt{\dfrac1{\sin^2(x)}-1}}=-\dfrac1{\sqrt{\csc^2(x)-1}}\,.\end{align}$
Since $\,y=\csc(x)\,,\,$ the function $\,\tan(x)\,$ is invertible on $\,\left(-\dfrac{\pi}2,\dfrac{\pi}2\right)\,$ and $\,x\in\left(-\dfrac{\pi}2,0\right)\,,\,$ it follows that
$x=\tan^{-1}\left(\!-\dfrac1{\sqrt{\csc^2(x)-1}}\!\right)=-\tan^{-1}\left(\!\dfrac1{\sqrt{y^2-1}}\!\right)\,.$
Consequently,
$\csc^{-1}(y)=-\tan^{-1}\left(\!\dfrac1{\sqrt{y^2-1}}\!\right)\quad$ for any $\;y\in\big(\!-\infty,-1\big)\,.$
Analogously, you can prove that
$\csc^{-1}(y)=\tan^{-1}\left(\!\dfrac1{\sqrt{y^2-1}}\!\right)\quad$ for any $\;y\in\big(1,+\infty\big)\,.$