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.
$\begin{align}-\dfrac{\sin t}{\cos t}\div (1-\sec^2t) &= -\dfrac{\sin t}{\cos t}\div \left(1-\dfrac{1}{\cos^2t}\right) = -\dfrac{\sin t}{\cos t}\div \left(\dfrac{\cos^2t-1}{\cos^2t}\right) \\&=-\dfrac{\sin t}{\cos t}\times \left(\dfrac{\cos^2t}{\cos^2t-1}\right)=\dfrac{\sin t \cos t}{1-\cos^2t} =\dfrac{\sin t \cos t}{\sin^2t} \\&=\dfrac{\cos t}{\sin t}\end{align}$