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# Please....I really need help with these pre-calculus hw questions posted below.?

1) Explain why arccos(12) has no solution.

2) Explain why someone would think sec(arcsec(0)) = 0

3) Explain why sec(arcsec(0)) actually has no solution.

4) arctan(θ) = 4/3 and the slope of m is -2. Find the equation of line m.

Please tell me how you get the answer and what the answer is. Thanks a lot. I will give 10 points to the best answer that covers the question in a detailed manner - by today.

How do I choose the best answer?- meaning which button do I press?

### 2 Answers

- LastOneStandingLv 51 decade agoFavorite Answer
1. The domain of the inverse cosine (arc cos) function is -1 to 1 (inclusive). 12 does not fall in this range. Therefore, arc cos (12) has no solution

2. I'd think that people would use the property that applies to sine and cosine:

sin ( arc sin (x)) = x

cos (arc cos (x)) = x

3. The graph of arc secant looks like this:

http://mathworld.wolfram.com/InverseSecant.html

It has an asymptote trailing y = 0. It never approaches it, so therefore, there is no solution for the problem

4. Rewrite the expression so its

tan θ = 4/3 (rather than inverse tangent)

Draw a triangle that satisfies this condition and solve for the third side. The triangle will lie in the first quadrant because the inverse tangent graph has y values that lie in the 1st and 2nd Quadrant and tangent is positive in the 1st quadrant only.

For this problem, is there a specific point that line m passes through? If so, it can be assumed it passes through the origin (a common point with the triangle) and so the line's equation would be y = -2x.

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