How to prove $\cot ^2x+\sec ^2x=\tan ^2x+\csc ^2x$? How can I prove the following equation?
\begin{eqnarray}
\cot ^2x+\sec ^2x &=& \tan ^2x+\csc ^2x\\
{{1}\over{\tan^2x}}+{{1}\over{\cos^2x}} &=& {{\sin^2x}\over{\cos^2x}}+{{1}\over{\sin^2x}}\\
{{\sin^2x+\cos^4x}\over{\sin^2x\cos^2x}} &=& {{\sin^4x+\cos^2x}\over{\sin^2x\ \cos^2x}}\\
\end{eqnarray}
Then, what can I do...?
Thank you for your attention.
 A: HINT:
We know, $$\csc^2x-\cot^2x=1=\sec^2x-\tan^2x$$
A: Well if nothing else comes to mind try by hand$$\cot^2 x+\sec^2x=\frac{\cos^2x}{\sin^2x}+\frac 1{\cos^2x}=\frac{\cos^4x+\sin^2x}{\cos^2x\sin^2x}$$
and $$\tan^2x+\csc^2x=\frac {\sin^2x}{\cos^2x}+\frac 1{\sin^2x}=\frac{\sin^4x+\cos^2x}{\cos^2x\sin^2x}$$
And these are equal if $$\cos^4x+\sin^2x=\sin^4x+\cos^2x$$
Now there are various ways to see it. Of course it is easier knowing the standard identities and using them, but they all pretty much boil down to $\sin^2x+\cos^2x=1$, which is in turn another way of writing Pythagoras, and which will definitely help here.
Note also that I've ignored any issue of infinite values or discontinuities. I have been careful not to divide by zero, which is quite easy to do by accident when working with trigonometric sums.
A: Hint: $1+\cot ^2x = \dfrac{\sin^2x+\cos^2x}{\sin^2x} = \csc ^2x$, $1+\tan ^2x = \dfrac{\cos^2x + \sin^2x}{\cos^2x} = \sec ^2x$.
A: Do you know, $\cot^2(x) + 1 = \csc^2(x)$ and $\tan^2(x) + 1 = \sec^2(x)$
So, $\cot ^2x+\sec ^2x=\csc^2(x)-1+\tan^2(x) + 1=\tan^2(x)+\csc^2(x)$
Thus, we end up with : $\cot ^2x+\sec ^2x=\tan^2(x)+\csc^2(x)$... 
