$\frac{1}{\sin^2(x)}+\frac{1}{\cos^2(x)} = \cos^2(x)+\sin^2(x) ?$

Question

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

Answer 1

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.

Answer 2

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 $b$.

Answer 3

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 $\cos^2\theta$.

Answer 4

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).