Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?

$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$


This is straightforward when you have another identity in your pocket, namely
$$\sec\theta \cdot \csc\theta = \tan\theta + \cot\theta \tag{$\star$}$$
This implies
$$\sec^2\theta \csc^2\theta = ( \tan\theta + \cot\theta )^2 = \tan^2\theta + 2 \tan\theta\cot\theta + \cot^2\theta = \tan^2\theta + 2 + \cot^2\theta$$
which gets you the equivalent of the identity in question. $\square$

By the way, proof of $(\star)$ arises from this trigonograph:

Computing the area of the big triangle in two ways gives
$$\frac{1}{2}\;\sec\theta\cdot\csc\theta = \frac{1}{2}\cdot 1 \cdot\left(\;\tan\theta + \cot\theta\;\right)$$
and the identity follows. $\square$
