Given $\cotθ = 1$ and angle θ lies in the third quadrant, determine the value of $\secθ$ and $\cscθ$

Question

Recently learned about the reciprocal trig ratios and I'm still don't fully understand them. Wondering if I can get a little help not only with this specific question but the calculating reciprocal trig ratios in general please and thank you.

Answer 1

For these types of problems, the hardest thing to do is finding what $\theta$ actually is. As for calculating these "other" trigonometric functions just use the definition, that is $$ \csc\theta = \frac{1}{\sin\theta}, \\ \sec\theta = \frac{1}{\cos\theta}, \\ \cot\theta = \frac{1}{\tan\theta}.$$

Back to your problem. By definition $$\cot\theta = \frac{1}{\tan\theta} = \frac{1}{\frac{\sin\theta}{\cos\theta}} = \frac{\cos\theta}{\sin\theta}.$$ You are given that $\cot\theta = 1$, which means that $\frac{\cos\theta}{\sin\theta} = 1$ and so $\cos\theta = \sin\theta$. Since $\theta$ lies in the third quadrand this implies that $\theta = \frac{5\pi}{4}$. I will calculate $\sec\theta$ for you: $$\sec\theta = \sec\frac{5\pi}{4} = \frac{1}{\cos\frac{5\pi}{4}} = \frac{1}{-\frac{\sqrt{2}}{2}} = -\frac{2}{\sqrt{2}} = -\sqrt{2}.$$ I will leave it to you to calculate $\csc\theta.$

Answer 2

I regard Blaze's answer as superior to the answer that I am giving, because he focuses on identifying the specific angle. This is the natural way of attacking the problem. The alternative approach is to compute $\cos(\theta)$ and $\sin(\theta)$ without identifying $(\theta)$. Once this is done, you can use the formulas $\sec(\theta) = \frac{1}{\cos(\theta)}, \csc(\theta) = \frac{1}{\sin(\theta)}.$

Alternative approach:

Since $\theta$ is in the 3rd quadrant, both $\cos(\theta)$ and $\sin(\theta)$ are negative.

Since $\frac{\cos(\theta)}{\sin(\theta)} = \cot(\theta) = 1$, $\cos(\theta) = \sin(\theta) \implies \cos^2(\theta) = \sin^2(\theta).$

You also have the general formula that $\sin^2(\theta) + \cos^2(\theta) = 1.$

Therefore, since $\cos^2(\theta) = \sin^2(\theta), 2\sin^2(\theta) = 1 \implies $

$\sin^2(\theta) = \frac{1}{2} \implies \sin(\theta) = \pm \frac{1}{\sqrt{2}}.$

Therefore, since it is known that $\sin(\theta)$ is negative,

$\sin(\theta)$ must be $- \frac{1}{\sqrt{2}}.$

Then, since it is known that $\cos(\theta) = \sin(\theta)$

$\cos(\theta)$ must also be $- \frac{1}{\sqrt{2}}.$

Answer 3

From Pythagoras' identity: $\:\cos^2\theta+\sin^2\theta=1$, are instantly deduced the relations

• $1+\tan^2\theta=\sec^2\theta$, whence $\;\sec^2\theta=\dfrac{\cot^2\theta+1}{\cot^2\theta}$
• $\cot^2\theta+1=\csc^2\theta$.

If $\cot\theta=1$, this means $\sec^2\theta=2=\csc^2\theta$.

Furthermore, in the third quadrant, $\sin\theta$ and $\cos\theta$, hence $\csc\theta$, and $\sec\theta$ are negative, so that $$\csc\theta=\sec\theta=-\sqrt{2^{\vphantom n}}.$$