Given tan 𝜃 = −2/3 in QIV, find cos 𝜃 and csc 𝜃 ? Provide a sketch and calculations to support your answer. Write the ratio in reduced fractional form with a rationalized denominator if necessary.

Question

Given tan 𝜃 = −2/3 in QIV, find cos 𝜃 and csc 𝜃 ? Provide a sketch and calculations to support your answer. Write the ratio in reduced fractional form with a rationalized denominator if necessary.

Answer 1

First, let's sketch the triangle in QIV where tan 𝜃 = -2/3.

In QIV, the adjacent side is positive and the opposite side is negative. Let's assume the adjacent side is 2 units and the opposite side is 3 units. This way, tan 𝜃 = opposite/adjacent = -3/2.

Now, let's find the hypotenuse using Pythagorean theorem:
hypotenuse^2 = adjacent^2 + opposite^2
hypotenuse^2 = 2^2 + 3^2
hypotenuse^2 = 4 + 9
hypotenuse^2 = 13
hypotenuse = √13

Now, we can find cos 𝜃 and csc 𝜃:
cos 𝜃 = adjacent/hypotenuse = 2/√13
csc 𝜃 = 1/sin 𝜃 = 1/(-3/√13) = -√13/3

So, cos 𝜃 = 2/√13 and csc 𝜃 = -√13/3.