# 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...?

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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.

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HINT:

We know, $$\csc^2x-\cot^2x=1=\sec^2x-\tan^2x$$

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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 : $\cot ^2x+\sec ^2x=\tan^2(x)+\csc^2(x)$...

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Hint: $1+\cot ^2x = \dfrac{\sin^2x+\cos^2x}{\sin^2x} = \csc ^2x$, $1+\tan ^2x = \dfrac{\cos^2x + \sin^2x}{\cos^2x} = \sec ^2x$.

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