$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$ 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
 A: The (multiplicative) inverse of $\sin^2\theta+\cos^2\theta$ is
$$\frac{1}{\sin^2\theta+\cos^2\theta},$$
and in general 
$$\frac{1}{a+b}\neq \frac{1}{a}+\frac{1}{b},$$
for 'numbers' $a$ and $b$.
A: No, since: $$\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}\neq \dfrac1{\cos^2\theta+\sin^2\theta}.$$
$$\dfrac{1}{\sin^2\theta}+\dfrac{1}{\cos^2\theta}=\dfrac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}=\dfrac{1}{\sin^2\theta\cos^2\theta}\neq1.$$
In general,  $\tfrac1a+\tfrac1b=\tfrac1{a+b}$ is not true, that's like saying that one half plus one half equals one quarter.
A: 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).
A: It is not true that the reciprocal of $a+b$ is the reciprocal of $a$ plus the reciprocal of $b$.  The reciprocal of $\sin^2\theta+\cos^2\theta$ is not the reciprocal of $\sin^2\theta$ plus the reciprocal of $\cos^2\theta$.
