<s>[INST] 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?

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

 [/INST] hint: $\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} = \dfrac{a^4+b^4}{a^2b^2}= \dfrac{(a^2+b^2)^2}{a^2b^2}-2$. Apply this identity for $a = \cos \theta, b = \sin \theta$ to get the desired identity.
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