Question

If sin theta = 8/17 and cot theta < 0, what is sec theta?

Answer 1

Answer:

$-\frac{17}{15}$

Step-by-step explanation:

By definition, $\cot \theta=\frac{1}{\tan \theta}$ and $\sec \theta=\frac{1}{\cos \theta}$. Since since $\cot \theta$ is negative, $\tan \theta$ must also be negative, and since $\sin \theta$ is positive, we must be in Quadrant II.

In a right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. The cosine of an angle in a right triangle is equal to its adjacent side divided by the hypotenuse. Therefore, we can draw a right triangle in Quadrant II, where the opposite side to angle theta is 8 and the hypotenuse of the triangle is 17.

To find the remaining leg, use to the Pythagorean Theorem, where $a^2+b^2=c^2$, where $c$ is the hypotenuse, or longest side, of the right triangle and $a$ and $b$ are the two legs of the right triangle.

Solving, we get:

$8^2+b^2=17^2,\\b^2=17^2-8^2,\\b^2=\sqrt{17^2-8^2}=\sqrt{225}=15$

Since all values of cosine theta are negative in Quadrant II, all values of secant theta must also be negative in Quadrant II.

Thus, we have:

$\sec\theta=\frac{1}{\cos \theta}=-\frac{1}{\frac{15}{17}}=\boxed{-\frac{17}{15}}$