Is division by $1-\frac1{\cos^2t}$ the same as multiplication by $1-\cos^2t$? If I have
$$ -\frac{\sin t}{\cos t} \div (1 - \sec^2t)$$
how do I simplify the divisor?
Let's say that I rewrite it as $$1 - \sec^2t = 1 - \frac{1}{\cos^2t}$$
Can someone detail the exact algebraic steps there? My basic question is:

Can I just say the following?
$$ -\frac{\sin t}{\cos t} \div \left(1 - \frac{1}{\cos^2t}\right) =  -\frac{\sin t}{\cos t} \times (1 - \cos^2t)$$

I suppose somebody didn't teach me fraction algebra like this in middle school, and now I'm struggling on petty things like this.
 A: No.
$\begin{align}-\dfrac{\sin t}{\cos t}\div (1-\sec^2t) &= -\dfrac{\sin t}{\cos t}\div \left(1-\dfrac{1}{\cos^2t}\right) = -\dfrac{\sin t}{\cos t}\div \left(\dfrac{\cos^2t-1}{\cos^2t}\right) \\&=-\dfrac{\sin t}{\cos t}\times \left(\dfrac{\cos^2t}{\cos^2t-1}\right)=\dfrac{\sin t \cos t}{1-\cos^2t} =\dfrac{\sin t \cos t}{\sin^2t} \\&=\dfrac{\cos t}{\sin t}\end{align}$
A: Let $$M=1-\frac{1}{\cos^22t}$$ and $$N=1-\cos^22t.$$ The question is asking whether $M$ and $N$ are multiplicative inverses, or in other words whether $MN=1$ in most interesting cases. Let's check. We find that $MN$ is equal to
$$\left(1-\frac{1}{\cos^22t}\right)(1-\cos^22t)=2-\left(\cos^22t+\frac{1}{\cos^22t}\right),$$ which is clearly not identically equal to $1,$ which puts the answer to your question in the negative.
A: $$ - \frac{ \sin{t} }{ \cos{t} } \div (1 - \frac{1}{\cos^2{t}}) = \\ - \frac{ \sin{t} }{ \cos{t} } \div (\frac{\cos^2{t} - 1}{\cos^2{t}})  = \\ - \frac{ \sin{t} }{ \cos{t} } \div (\frac{- \sin^2{t}}{\cos^2{t}}) = \\- \frac{ \sin{t} }{ \cos{t} } \cdot (\frac{- \cos^2{t}}{\sin^2{t}})  = \\ =\frac{\cos{t}}{\sin{t}}$$
Using the fact that: $$-\sin^2{t} = \cos^2{t}  - 1$$
A: Note that
$$1-\frac1x = 1-x \iff \frac1x=x  \iff x^2=1$$
that is by $x=\cos^2t$
$$\cos^4t=1 \iff t=k\pi$$
which is not true in general and therefore your step is not correct.
A: No.
$\lvert \sec x\rvert \ge 1$, so $1-\sec^2 x \le 0$, so division by $1-\sec^2 x$ flips the sign (if the division is defined).
$0 \le \lvert \cos x\rvert \le 1$, so $0\le 1 - \cos ^2 x \le 1$, so multiplication by $1-\cos ^2 x$ keeps the sign unchanged.
