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?

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

 A: \begin{align*}
\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} &= \tan^2\theta + \cot^2\theta \\
&= \sec^2\theta - 1+ \csc^2\theta - 1\\
&= \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} - 2\\
&= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} -2\\
&= \sec^2\theta \csc^2\theta - 2
\end{align*}
A: 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.
A: First, let's make a common denominator of $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}$.
Camera? Action!
$$\require{cancel}\begin{aligned}\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta}&=\frac{\sin^4{\theta}+\cos^4{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{\left(\sin^2{\theta}+\cos^2{\theta}\right)^2-2\sin^2{\theta}\cos^2{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{1-2\sin^2{\theta}\cos^2{\theta}}{\sin^2{\theta}\cos^2{\theta}}\\&=\frac{1}{\sin^2{\theta}\cos^2{\theta}}-\frac{2\cancel{\sin^2{\theta}\cos^2{\theta}}}{\cancel{\sin^2{\theta}\cos^2{\theta}}}\\&=\frac{1}{\sin^2{\theta}\cos^2{\theta}}-2\\&=\csc^2{\theta}\sec^2{\theta}-2\\&=\sec^2{\theta}\csc^2{\theta}-2\end{aligned}$$
A: hint: $a^4+b^4 = a^4+2a^2b^2+b^4-2a^2b^2=(a^2+b^2)^2 - 2a^2b^2$
A: $$\frac{s^2}{c^2}+\frac{c^2}{s^2}=\frac{s^4+c^4}{s^2c^2}=\frac{(s^2+c^2)^2-2s^2c^2}{s^2c^2}.$$
A: 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. $\square$
