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.
No.
$\begin{align}-\dfrac{\sin t}{\cos t}\div (1-\sec^2t) &amp;= -\dfrac{\sin t}{\cos t}\div \left(1-\dfrac{1}{\cos^2t}\right) = -\dfrac{\sin t}{\cos t}\div \left(\dfrac{\cos^2t-1}{\cos^2t}\right) \\&amp;=-\dfrac{\sin t}{\cos t}\times \left(\dfrac{\cos^2t}{\cos^2t-1}\right)=\dfrac{\sin t \cos t}{1-\cos^2t} =\dfrac{\sin t \cos t}{\sin^2t} \\&amp;=\dfrac{\cos t}{\sin t}\end{align}$
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.