<s>[INST] $\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ This is a simple question.  Since $\cos(\theta)^2 + \sin(\theta)^2 = 1$
Can I take the inverse of this $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \frac{1}{1}$?
Finally getting $\frac{1}{\cos^2(\theta)}+\frac{1}{\sin^2(\theta)} = \cos(\theta)^2 + \sin(\theta)^2$
If this is incorrect thinking can someone please put me on the right path?
Thanks
 [/INST] Set $\theta=\pi/4$; then $\sin\theta=\cos\theta=1/\sqrt{2}$; therefore
$$
\frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta}=
\frac{1}{1/2}+\frac{1}{1/2}=2+2=4\ne1.
$$
We might ask when equality holds:
$$
\frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta}=1
$$
is equivalent to
$$
\frac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}=1
$$
which becomes
$$
\sin^2\theta\cos^2\theta=1.
$$
We can multiply both sides by $4$ getting
$$
4\sin^2\theta\cos^2\theta=4.
$$
or
$$
\sin^2(2\theta)=4
$$
which has no solution. So, not only $\theta=\pi/4$ is a counterexample, but all values of $\theta$ are (excluding integer multiples of $\pi/2$ that make the left hand side undefined).
</s>