Evaluate tangent left parenthesis inverse cosine a right parenthesis.

Question

(1 point)
Responses

Start Fraction Start Root 1 minus a squared End Root over 1 minus a squared End Fraction
Image with alt text: Start Fraction Start Root 1 minus a squared End Root over 1 minus a squared End Fraction

Start Fraction Start Root 1 minus a squared End Root over a End Fraction
Image with alt text: Start Fraction Start Root 1 minus a squared End Root over a End Fraction

Start Root 1 minus a squared End Root
Image with alt text: Start Root 1 minus a squared End Root

Start Fraction a Start Root 1 minus a squared End Root over 1 minus a squared End Fraction

Answer 1

To evaluate $\tan ({\cos }^{-1}(a))$, let's start by letting $\theta ={\cos }^{-1}(a)$. This means that $\cos (\theta )=a$.

From the Pythagorean identity, we know that:

$${\sin }^2(\theta )+{\cos }^2(\theta )=1$$

Substituting $\cos (\theta )=a$:

$${\sin }^2(\theta )+a^2=1$$

Thus, we can express ${\sin }^2(\theta )$ as:

$${\sin }^2(\theta )=1-a^2$$

Taking the square root, we find:

$$\sin (\theta )=\sqrt{1-a^2}$$

Now we can compute $\tan (\theta )$:

$$\tan (\theta )=\frac{\sin (\theta )}{\cos (\theta )}=\frac{\sqrt{1-a^2}}{a}$$

Therefore, the value of $\tan ({\cos }^{-1}(a))$ is:

$$\frac{\sqrt{1-a^2}}{a}$$

The correct response is:

$$\text{Start Fraction Start Root 1 minus a squared End Root over a End Fraction}$$