<s>[INST] A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere. 
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of the equation.
Work so far:
RS:
$$
\begin{align}
& \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \\[10pt]
& = \frac{\sin^4\theta}{(\cos^2\theta)(\sin^2\theta)} + \frac{\cos^4\theta}{(\sin^2\theta) (\cos^2\theta)} + \frac{(\sin^4\theta)(\cos^2\theta)}{(\sin^2\theta)(\cos^2\theta)} + \frac{(\sin^2\theta)(\cos^4\theta)}{(\sin^2\theta)(\cos^2\theta)} \\[10pt]
& = \frac{\sin^4\theta + \cos^4\theta + (\sin^4\theta)(\cos^2\theta) + (\sin^2\theta)(\cos^4\theta)}{(\cos^2\theta)(\sin^2\theta)}
\end{align}
$$
I am completely lost after this.
 [/INST] Look at the largest triangle.

(There's a reason my avatar --the logo of my software company-- is a stylized version of this figure.)
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