Question

NO LINKS!! Please help me with this problem. Part 2​

Answer 1

B is the correct option.

Using the definition of each function as well as the Pythagorean Theorem, these values are found. :)

Answer 2

Answer: Choice B

`-(√(91))/(3)`

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Step-by-step explanation:

We're given
`\cos(\theta) = -(3)/(10)`

Use that value of cosine to find sine.

`\sin^2(\theta) + \cos^2(\theta) = 1\\\\\sin^2(\theta) = 1-\cos^2(\theta)\\\\\sin(\theta) = √(1-\cos^2(\theta)) \ \text{ ... sine is positive in Q2}\\\\\sin(\theta) = \sqrt{1-\left(-(3)/(10)\right)^2}\\\\\sin(\theta) = \sqrt{1-(9)/(100)}\\\\\sin(\theta) = \sqrt{(100)/(100)-(9)/(100)}\\\\\sin(\theta) = \sqrt{(100-9)/(100)}\\\\\sin(\theta) = \sqrt{(91)/(100)}\\\\\sin(\theta) = (√(91))/(√(100))\\\\\sin(\theta) = (√(91))/(10)\\\\`

Now divide the sine and cosine values to find tangent.

`\tan(\theta) = (\sin(\theta))/(\cos(\theta))\\\\\tan(\theta) = \sin(\theta) / \cos(\theta)\\\\\tan(\theta) = (√(91))/(10) / -(3)/(10)\\\\\tan(\theta) = (√(91))/(10) * -(10)/(3)\\\\\tan(\theta) = -(√(91))/(3)\\\\`

Tangent is always negative in Q2.