find the exact value of $\sin(\tan^{-1}(-2))$$\bigcirc\ -\dfrac{1}{2}$$\bigcirc\ -\dfrac{2}{\sqrt{5}}$$\bigcirc\ \dfrac{2}{\sqrt{5}}$$\bigcirc\ \dfrac{1}{\sqrt{5}}$$\bigcirc\ -\dfrac{1}{\sqrt{5}}$

Question

Accepted Answer

# Explanation:
## Step1: Let $ \theta = \tan^{-1}(-2) $, so $ \tan\theta = -2 $ and $ \theta $ is in the fourth quadrant (since tangent is negative).
We know that $ \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{-2}{1} $ (we take the absolute values for the sides of the right triangle).
## Step2: Find the hypotenuse $ r $ using the Pythagorean theorem: $ r = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} $.
## Step3: Now, $ \sin\theta = \frac{\text{opposite}}{r} $. Since $ \theta $ is in the fourth quadrant, sine is negative. So $ \sin(\tan^{-1}(-2)) = \sin\theta = \frac{-2}{\sqrt{5}} $.
# Answer:
$ -\dfrac{2}{\sqrt{5}} $ (corresponding to the option: $ -\dfrac{2}{\sqrt{5}} $)

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Question

Explanation:

Step1: Let \( \theta = \tan^{-1}(-2) \), so \( \tan\theta = -2 \) and \( \theta \) is in the fourth quadrant (since tangent is negative).

Step2: Find the hypotenuse \( r \) using the Pythagorean theorem: \( r = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \).

Step3: Now, \( \sin\theta = \frac{\text{opposite}}{r} \). Since \( \theta \) is in the fourth quadrant, sine is negative. So \( \sin(\tan^{-1}(-2)) = \sin\theta = \frac{-2}{\sqrt{5}} \).

Answer: