Question

Please help me. I need help

Answer 1

Look at the first picture.

In Quadrant II tangent is negative and cosine is negative too.

We have:

`\tan\theta=-(2)/(3)=(2)/(-3)`

therefore we have the point (-3, 2) → x = -3 and y = 2.

Calculate r:

`r=√((-3)^2+2^2)=√(9+4)=√(13)`

Calculate cosine:

`\cos\theta=(x)/(r)\to\cos\theta=(-3)/(√(13))\\\\\cos\theta=-(3)/(√(13))\cdot(√(13))/(√(13))\\\\\cos\theta=-(3√(13))/(13)`

Your answer is
`\boxed{a.\ -(3√(13))/(13)}`

Answer 2

a. -(3√13)/13

Explanation:

The cosine can be found from the tangent by way of the secant.

tan(θ)² +1 = sec(θ)² = 1/cos(θ)²

Then ...

cos(θ) = ±1/√(tan(θ)² +1)

The cosine is negative in the second quadrant, so we will choose that sign.

cos(θ) = -1/√((-2/3)² +1) = -1/√(4/9 +1) = -1/√(13/9)

cos(θ) = -3/√13 = -(3√13)/13 . . . . . matches your selection A