I was working on trig homework and came across a question that I didn't understand how to even begin to approach. It asked us to use trigonometric identities to write $\csc(x)$ in terms of $\sec(x)$.
I'm not sure what I can do. I had though the cofunction identity $\sec \left(\frac{\pi}{2} – x \right) = \csc x$ was the answer, but I was told I was wrong, with no explanation. This identity was on a handout provided by the teacher.
How would I go about solving this problem and problems like it? Also, why was my initial answer wrong? Any help would be greatly appreciated.
Best Answer
It’s wrong because you didn’t write $\csc x$ in terms of $\sec x$: $\frac{\pi}2-x$ isn’t $x$. I suspect that you were intended to do something like this:
$$\csc x=\frac1{\sin x}\;,$$ and $$\sec x=\frac1{\cos x}\;,$$ so
$$\begin{align*}&\frac1{\csc^2 x}+\frac1{\sec^2x}=1\;,\\ &\frac1{\csc^2x}=1-\frac1{\sec^2x}=\frac{\sec^2x-1}{\sec^2x}\;,\\ &\csc^2x=\frac{\sec^2x}{\sec^2x-1}\;,\\ &\csc x=\pm\frac{\sec x}{\sqrt{\sec^2x-1}}\;. \end{align*}$$
For a complete answer you still have to sort out where $\csc x$ is positive and where it’s negative, which is just a bit messy.