The question I'm given is: "if $\csc\theta = -17/15$ and $\theta$ is in the third quadrant, determine all possible values of the expression $\tan\theta + 3\sec \theta$. State all possible values of $\theta$ in radians."

I know to flip $-17/15$ to make it $\sin\theta$ but not sure what to do from there?

  • 1
    $\begingroup$ If a triangle has a hypotenuse of 17 and on opposite side of 15, what would the adjacent side be? $\endgroup$
    – John Joy
    May 5, 2016 at 23:25

1 Answer 1


I will outline a method for solving the problem:

You found that $\sin\theta = -\dfrac{15}{17}$.

You can use the Pythagorean Identity $\sin^2\theta + \cos^2\theta = 1$ to solve for $\cos\theta$. $$\cos^2\theta = 1 - \sin^2\theta$$ Since $\theta$ is a third-quadrant angle, $\cos\theta < 0$. Thus, $$\cos\theta = -\sqrt{1 - \sin^2\theta}$$ Once you calculate $\cos\theta$, you can find $\tan\theta$ and $\sec\theta$ using the identities \begin{align*} \tan\theta & = \frac{\sin\theta}{\cos\theta}\\ \sec\theta & = \frac{1}{\cos\theta} \end{align*} Substitute the values you find into the expression $\tan\theta + 3\sec\theta$.

Alternatively, you could follow the suggestion that John Joy made in the comments. Use the Pythagorean Theorem to determine $|x|$ in the diagram below. Since $x$ is in the third quadrant, $x < 0$. From there, you can use the triangle to find the values of the trigonometric functions of $\theta$.

third-quadrant triangle

Since both sine and cosine are negative in the third quadrant, the tangent is positive. Since the arctangent function has range $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ and $\tan\theta > 0$, $$0 < \arctan(\tan\theta) < \frac{\pi}{2}$$ Since $\theta$ is a third-quadrant angle, $\arctan(\tan\theta)$ is a first-quadrant angle, and tangent has period $\pi$, one possible value for $\theta$ is $$\pi + \arctan(\tan\theta)$$ The other possible values of $\theta$ are found by adding integral multiples of $2\pi$ to $\pi + \arctan(\tan\theta)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .