Question

cotx=-3/5,-pi/2<x<0
Use identities to find the exact values for the remaining five trigonometric functions

Answer 1

We use the fact that $\cot x = \frac{1}{\tan x}$ and $\tan^2 x + 1 = \sec^2 x$.

Since $\cot x = -\frac{3}{5}$ and $-\frac{\pi}{2} < x < 0$, we know that $\tan x = -\frac{5}{3}$ (since $\tan x$ is negative in the second quadrant). Thus, we have

$$\sec^2 x = \tan^2 x + 1 = \left(-\frac{5}{3}\right)^2 + 1 = \frac{34}{9}$$

Taking the square root, we get $\sec x = \pm \frac{\sqrt{34}}{3}$. Since $-\frac{\pi}{2} < x < 0$, we know that $\cos x < 0$, so $\sec x$ is negative. Therefore, we have $\sec x = -\frac{\sqrt{34}}{3}$.

Now we can find the other trigonometric functions:

$$\begin{aligned} \sin x &= \frac{\tan x}{\sec x} = -\frac{5}{\sqrt{34}} \\ \cos x &= -\frac{1}{\sec x} = \frac{\sqrt{34}}{3} \\ \csc x &= \frac{1}{\sin x} = -\frac{\sqrt{34}}{5} \\ \sec x &= -\frac{\sqrt{34}}{3} \\ \cot x &= -\frac{3}{5} \end{aligned}$$

Therefore, the exact values of the six trigonometric functions are:

$$\begin{aligned} \sin x &= -\frac{5}{\sqrt{34}}, \\ \cos x &= \frac{\sqrt{34}}{3}, \\ \tan x &= -\frac{5}{3}, \\ \csc x &= -\frac{\sqrt{34}}{5}, \\ \sec x &= -\frac{\sqrt{34}}{3}, \\ \cot x &= -\frac{3}{5}. \end{aligned}$$