Question

If `cot A = -8/5`, and `sin A<=0` what is `(cos A)^2-4sinA`?

Answer 1

First of all, `cotA = 1/tanA`. Tangent is negative in quadrants `II` and `IV`. Sine is negative in quadrants `III` and `IV`. Hence, we can conclude `A` is in quadrant `IV`.

We know that `cottheta = 1/tantheta = 1/("opposite"/"adjacent") = "adjacent"/"opposite"`

So, we know that the side adjacent `A` measures `8` and the side opposite measures `-5`.

To find `cosA` and `sinA`, we need to determine the length of the hypotenuse. By pythagorean theorem:

`(8)^2 + (-5)^2 = h^2`

`64 + 25 = h^2`

`89 = h^2`

`h = sqrt(89)`

We define `cosA` as `"adjacent"/"hypotenuse"` and `sinA` as `"opposite"/"hypotenuse"`. So, `cosA = 8/sqrt(89)` and `sinA = -5/sqrt(89)`.

We now evaluate the expression accordingly.

`(cosA)^2 - 4sinA = (8/sqrt(89))^2 - 4(-5/sqrt(89)) = 64/89 + 20/sqrt89`.

Rationalize the denominator:

`=>64/89 + (20sqrt(89))/89 = (64 + 20sqrt(89))/89`

Hopefully this helps!

Answer 2

`(cosA)^2-4sinA=(64+20sqrt89)/89`

Explanation:

As `cotA=-8/5` and `sinA < 0` means both are negative and hence they lie in fourth quadrant. Hence, while `cosA` is positive and `sinA` is obviously negative. Calculate them

As `cotA=-8/5`, `csc^2A=1+(-8/5)^2=1+64/25=89/25` and `cscA=-sqrt89/5`

Hence, `sinA=-5/sqrt89` and as

`cosA=cotAxxsinA=-8/5xx-5/sqrt89=8/sqrt89` and

`(cosA)^2-4sinA=64/89-4xx-5/sqrt89=(64+20sqrt89)/89`