Trigonometry problem involving sine and tangent

Question

I'm struggling with a trigonometry problem and I was hoping someone could help me understand it better. The problem is as follows:

If $\sin \alpha=p$ and $\pi<\alpha<2 \pi$, what is $|\tan \alpha|$ equal to?

The options are:

(a) $\dfrac{p}{\sqrt{1-p^2}}$;

(b) $-\dfrac{p}{\sqrt{1-p^2}}$;

(c) $\pm \dfrac{p}{\sqrt{1-p^2}}$;

(d) none of the above.

I tried to solve this problem by using the identity $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$ and then finding $\cos \alpha$ using the Pythagorean identity. This gave me:

$$ \cos \alpha = \sqrt{1-\sin^2 \alpha} = \sqrt{1-p^2} $$

So,

$$ |\tan \alpha| = \frac{|\sin \alpha|}{|\cos \alpha|} = \frac{p}{\sqrt{1-p^2}}$$

However, according to the answer key, the correct answer is (b). I don't understand why the absolute value of $\tan \alpha$ should be negative. Shouldn't it be positive since the absolute value of $\sin \alpha$ is positive and $\cos \alpha$ is negative in the third quadrant where $\alpha$ lies?

I would appreciate any help in understanding this problem better.

Thank you in advance.

Answer 1

Since $\pi<\alpha<2\pi$, $p(=\sin\alpha)<0$. Therefore $$ |\tan\alpha|=\frac{|\sin\alpha|}{|\cos\alpha|}=\frac{-p}{\sqrt{1-p^2}}. $$

Answer 2

The answer to this query is hidden in the problem itself!

It seems like you are applying Textbook Mathematics to a problem which should be treated geometrically.

Given that $$ \pi < \alpha < 2\pi,$$ meaning $\alpha$ is an obtuse angle and lies in third or fourth quadrant. It would be convenient to let $$\beta = \alpha - \pi$$ $$\Rightarrow sin\,\alpha = -sin\,\beta$$

Let's comprehend the equation: $$ sin \,\alpha = p$$

Here is the geometric representation of the given problem.

$\boldsymbol \rightarrow$ Geometric solution

I have taken a Unit circle (circle of radius 1 unit). Since $sin \,\alpha = \frac{p}{1}$ , the hypotenuse and the side opposite to $\alpha$ is known. Using Pythagoras theorem, we find that the third side is equal to $\sqrt{{1} - {p^2}}$.

$\star \,$ Since $0<\beta <\pi$ , $sin \,\beta$ is always positive. This in turn means that $sin\,\alpha$ is always negative.

But $sin\,\alpha = p$

$\Rightarrow \text{p is negative.}$

Now, $\qquad$ $tan\,\beta = \frac{p}{\sqrt{{1} - {p^2}}}$

$\star \,$ Note that we used p without considering its sign. This is because $tan\,\beta$ can be positive or negative depending on quadrant. This means the sign will take care of itself!

Using the relation $tan\,\alpha = - tan\,\beta$, we find $$tan\,\alpha = -\frac{p}{\sqrt{{1} - {p^2}}}$$