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:

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