Expand this proof further: $\cot^2\theta + \sec^2\theta = \tan^2\theta + \csc^2\theta$

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

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

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

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

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

• Laughing at myself now, I pretty much derived that earlier. Maybe I should sleep more. Thank you. – Kermit the Hermit Jun 16 '14 at 5:32

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.

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

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

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

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

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