Let $x = a\sec\theta$ where $a>0$, $0<\theta<\frac{\pi}{2}$ or $\pi < \theta < \frac{3\pi}{2}$. Find the value of $\tan\theta$.

My attempt:

I want to use a geometric method to solve this question since I know how to do it via trigonometric identities. I use the following definition provided by my book:

For a general angle $\theta$, we let $P(x,y)$ be any point on the terminal side of $\theta$ and we let $r$ be the distance $|OP|$, then we define:

enter image description here

So we know $\cos\theta = \frac{a}{x}$ and for $0<\theta<\frac{\pi}{2}$, $x=a\sec\theta$ is clearly positive. So, using my book's definition, the point $P(x,y) = P(a, \sqrt{x^2-a^2})$, so $\tan\theta = \frac{\sqrt{x^2-a^2}}{a}$.

Now where I have the problem is for $\pi < \theta < \frac{3\pi}{2}$. When $\pi < \theta < \frac{3\pi}{2}$, $x = a\sec\theta<0$ and $a>0$ as per the question. But, according to my book's definition, we should be in the third quadrant for $\pi < \theta < \frac{3\pi}{2}$, but $\cos\theta=\frac{a}{x}$ doesn't make sense in the third quadrant because if you draw a right angle triangle in the third quadrant, $a$ (the base of the triangle) should be negative (i.e., have a negative $x$-coordinate) and $x$ (the hypotenuse) should be positive, but we have $a>0$ and $x<0$. Could someone clear up my understanding?

  • $\begingroup$ Which software/website have you used make that figure? $\endgroup$ – sgrmshrsm7 Mar 17 '17 at 20:21

Things are a little confusing since $x$ is both a coordinate in the figure you're using to define the $\sec$ function and also a variable in the problem you're trying to solve. So I'll write $t$ for the horizontal coordinate in the figure and formulas for the trigonometric functions, and therefore $$ \sec \theta = \frac rt \quad\text{and}\quad \tan \theta = \frac yt. $$

Now it appears that what you have done is, you have observed that when $0<\theta<\frac\pi2,$ so that the ray at angle $\theta$ lies in the first quadrant, there is a right triangle with leg $a$ adjacent to an angle of measure $\theta$ that "fits" perfectly between the ray and the horizontal axis, and the length of the hypotenuse of that triangle is $x.$

Now you expect to find a triangle whose side lengths you can similarly identify that fits in the same way between the horizontal axis and the ray at angle $\theta$ when $\pi < \theta < \frac{3\pi}{2}.$ There is, in fact, a suitable right triangle that has a hypotenuse of length $\lvert x \rvert > 0.$ There is just not a triangle whose hypotenuse you can naïvely label $x,$ expecting that label to makes sense as a length.

The variables $t$ and $y$ that show up in the definition of the trigonometric functions for general angles are not lengths of the sides of the triangle; they are coordinates of the other end of the hypotenuse, away from the origin. Unlike the sides of a triangle, these coordinates can be negative.

When $\pi < \theta < \frac{3\pi}{2},$ both $t < 0$ and $y < 0,$ since $(t,y)$ is the coordinates of a point in the third quadrant. But we always have $r > 0,$ that is, $r$ is actually the distance between the points $(0,0)$ and $(t,y)$. Therefore $\sec\theta = \frac rt < 0,$ and therefore $x = a\sec\theta < 0$ since $a > 0.$

You can, if you like, construct a right triangle in the third quadrant with vertices at $(0,0),$ $(-a,0),$ $(-a,y),$ choosing $y$ so that the point $(-a,y)$ is on the ray at angle $\theta.$ Then the leg of that triangle along the horizontal axis will have length $a$ (the distance between $(0,0)$ and $(-a,0)$; observe that it is perfectly sensible to have a positive distance to a point that happens to have a negative coordinate). The hypotenuse of that triangle will have length $-x.$ The length is not $x$; the definition of $\sec\theta$ for a general angle $\theta$ never says that $a\sec\theta$ will be a length when $a$ is a length.

That's about as close as you can get to an accurate labeling of a triangle in the third quadrant comparable to the triangle you can construct in the first quadrant.

Yes, it can be confusing to know when the hypotenuse of a triangle should be said to have length $x$ and when it should have length $-x.$ I think that is a big part of the reason why the figure and formulas you found define the trigonometric functions of a general angle in terms of Cartesian coordinates instead of drawing a triangle.


Rewritng $-x = (-a)\sec \theta$ should lead you in the direction you want.

However, it is a bit confusing to write $P(x,y)$ when the letter $x$ already stands for something other than the first coordinate of $P$, so rather than write $P(x,y)$ instead I'll write $P(\tilde{a},\tilde{b})$ for the point on the terminal side of the angle $\theta$.

Fix $\pi < \theta < \frac{3\pi}{2}$. Clearly you should have that $\tilde{a},\tilde{b} < 0$.

What if we allowed $\tilde{a} := -a < 0$? Given that our angle $\theta$ is fixed, picking $\tilde{a}$ should determine both $\tilde{b},r$ uniquely. Note that $x = a \sec \theta < 0$ implies that $$ -x = (-a) \sec \theta > 0 \quad \mathrm{and} \quad \frac{\tilde{a}}{-x} = \frac{-a}{-x} = \cos \theta = \frac{\tilde{a}}{r}.$$

Therefore $r = -x > 0$, and furthermore, $\tilde{b} = -\sqrt{r^2 - \tilde{a}^2} = - \sqrt{x^2 - a^2}<0$ (you take the negative square root because $P$ is in the third quadrant).

Thus we conclude $$ \tan \theta = \frac{\tilde{b}}{\tilde{a}} = \frac{-\sqrt{x^2 - a^2}}{-a} = \frac{\sqrt{x^2 - a^2}}{a}$$


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