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I am curious as to how one would prove this without changing both sides of the equation. In other words going straight from $$\cot^2\theta + \sec^2\theta$$ to $$\tan^2\theta + \csc^2\theta$$ (or vice-versa) rather than editing both sides to meet up at a certain point as I have done in my proof.

This is my proof:

$$\cot^2\theta + \sec^2\theta = \tan^2\theta + \csc^2\theta$$

LHS: $$=\frac{\cos^2\theta}{\sin^2 \theta} +\frac{1}{\cos^2\theta}$$

$$=\frac{\cos^4\theta + \sin^2\theta}{ \sin^2\theta \cos^2\theta}$$

RHS: $$=\frac{\sin^2\theta}{\cos^2\theta} + \frac{1}{\sin^2\theta}$$

$$= \frac{{(\sin^2\theta)}^2 + \cos^2\theta}{\sin^2\theta\cos^2\theta}$$

$$=\frac{(1-\cos^2\theta)^2 + \cos^2\theta}{\sin^2\theta\cos^2\theta}$$

$$=\frac{\cos^4\theta + \sin^2\theta}{\sin^2\theta\cos^2\theta}$$

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$$\begin{align} \cot^2\theta + \sec^2\theta &=(\csc^2{\theta}-1)+(1+\tan^2{\theta})\\ &=\csc^2{\theta}+\tan^2{\theta} \end{align}$$

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    $\begingroup$ Laughing at myself now, I pretty much derived that earlier. Maybe I should sleep more. Thank you. $\endgroup$ – Kermit the Hermit Jun 16 '14 at 5:32
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Take your sequence of expressions labelled "RHS", flip it over and append it to your sequence of expressions labelled "LHS". Delete the duplicate at the join.

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Take difference of left hand side and right hand side of known identities and transpose:

$ \sec^2\theta -\tan^2\theta = 1 $

$ \csc^2\theta -\cot^2\theta = 1 $

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Hint: Use the fact $\tan^2{\theta}=\sec^2{\theta-1}$ and $\cot^2{\theta} = \csc^2{\theta}-1 $.

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$$\csc^2 A-\cot^2A=1=\sec^2A-\tan^2A$$

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