Questions: Evaluate. tan[cos^(-1)(-3/5)] a. -4/3 b. -3/4 c. 3/4 d. 4/3

Transcript text: Evaluate. \[ \tan \left[\cos ^{-1}\left(-\frac{3}{5}\right)\right] \] a. $-\frac{4}{3}$ b. $-\frac{3}{4}$ c. $\frac{3}{4}$ d. $\frac{4}{3}$

Solution

Solution Steps

To solve $\tan \left[\cos ^{-1}\left(-\frac{3}{5}\right)\right]$, we need to find the tangent of the angle whose cosine is $-\frac{3}{5}$. We can use the Pythagorean identity to find the sine of this angle, and then use the definition of tangent as the ratio of sine to cosine.

Let $\theta = \cos^{-1}\left(-\frac{3}{5}\right)$. Then, $\cos(\theta) = -\frac{3}{5}$.
Use the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$.
Solve for $\sin(\theta)$.
Compute $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$.

Step 1: Determine $\cos(\theta)$

We are given that \[ \cos(\theta) = -\frac{3}{5} = -0.6. \]

Step 2: Use the Pythagorean Identity

Using the Pythagorean identity \[ \sin^2(\theta) + \cos^2(\theta) = 1, \] we can find $\sin(\theta)$: \[ \sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}. \] Thus, \[ \sin(\theta) = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8. \] Since $\theta$ is in the second quadrant, $\sin(\theta)$ is positive.

Step 3: Calculate $\tan(\theta)$

Now, we can calculate $\tan(\theta)$ using the definition: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \approx -1.3333. \]