# Given that $\sec \theta = k$, $|k| \geq 1$, and that $\theta$ is obtuse, express in terms of $k$: $\cos \theta$, and $\csc \theta$

Given that $\sec \theta = k$, $|k| \geq 1$, and that $\theta$ is obtuse, express in terms of $k$: $\cos \theta$, and $\csc \theta$

For $\cos \theta$ I get: $\frac{-1}{k}, (as \theta$ is obtuse, however this is wrong too)

For $\csc \theta$ I get the answer to be: $\frac{k}{\sqrt{k^2 - 1}}$, (this is a wrong answer too).

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How do you know it's wrong? –  Gerry Myerson Mar 1 at 12:02
Answer in the back of the book –  Assad Mar 1 at 12:34
Yes, and what is the answer in the back of the book? –  Gerry Myerson Mar 1 at 12:37
cscθ=-k/√(k^2−1) –  Assad Mar 1 at 12:41
Ah. $\theta$ is obtuse, so $\sec\theta$ is negative, so $k$ is negative, but $\csc\theta$ is positive, so you need that $-k$. –  Gerry Myerson Mar 1 at 23:32
The first one is not difficult. By definition $$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{k}.$$
If by obtuse you mean $\pi / 2\leq \theta \leq \pi$ modulo $2\pi$ as usual, this implies $\cos\theta =k\leq 0$ and $\sin\theta \geq 0$.
Now $$\sin^2\theta+\cos^2\theta=1\quad\Rightarrow\quad|\sin\theta|=\sqrt{1-\cos^2\theta}=\sqrt{1-1/k^2}=\frac{\sqrt{k^2-1}}{|k|}.$$ Since $\sin\theta\geq 0$, you get $$\sin\theta = \frac{\sqrt{k^2-1}}{-k}\quad\Rightarrow\quad \csc\theta=-\frac{k}{\sqrt{k^2-1}}.$$