Is the denominator never zero in sec, cosec and cot functions?

Question

I am asking this question because i was told the following:

Since $\sec(y)=1/\cos(y)$ we know that $\cos(y)≠0$ and similarly for $\rm{cosec}$. Then you can rewrite like this: $3\sec(y)=4\csc(y)⟺3/\cos(y)=4/\sin(y).$

Answer 1

Yes, you can write it like that, provided the denominators are never $0$.

Answer 2

I think there's something subtle going on here. NO denominator can be zero. But we would sometimes ask, "When is the denominator zero," which is really a way of asking "where is the function not defined."

The function $\sec x$ doesn't have denominator, but it is undefined at $x=\pi/2$. The function $1/\cos x$ has a denominator. It's fair to say that the denominator is zero at $x=\pi/2$ and because of that, we're not going to let $x$ be $\pi/2.$

So when you're solving your equation, you're looking for values that make the equation true. You can throw out the values $y=\pi/2 + 2k\pi$ because secant isn't defined there and so these values can't be part of the solution set. Now that you've thrown those values out, it's safe to replace $\sec y$ by $1/\cos y.$