Use the Pythagorean identities rather than reference triangles.

Find tanθ and cotθ if secθ = 3/2 and sinθ <0

Please help me , thanks so much

Ditto

Draw the triangle in QIV so that

r = 3
x = 2
y = -√5

tanθ = y/x = -√5/2
...

To find tanθ and cotθ, we can use the Pythagorean identities:

1. sec²θ = tan²θ + 1, and
2. cot²θ + 1 = csc²θ.

Given that secθ = 3/2 and sinθ < 0, we can solve for the values of tanθ and cotθ using the Pythagorean identities.

First, let's find the value of cosθ using the given value of secθ:

secθ = 1/cosθ = 3/2

Cross-multiplying, we get:
2 = 3cosθ

Dividing both sides by 3:
2/3 = cosθ

To find the value of sinθ, we can use the Pythagorean identity:
sin²θ + cos²θ = 1

Plugging in the value of cosθ we just found:
sin²θ + (2/3)² = 1

Expanding and rearranging:
sin²θ + 4/9 = 1

Subtracting 4/9 from both sides:
sin²θ = 1 - 4/9
sin²θ = 5/9

Since sinθ < 0, we know that θ is in the third quadrant, where sinθ < 0 and cosθ < 0.

Taking the square root of both sides to find sinθ:
sinθ = -√(5/9)
sinθ = -√5/3

Now let's find tanθ using the Pythagorean identity:

sec²θ = tan²θ + 1

Plugging in the value of secθ we found earlier:
(3/2)² = tan²θ + 1

Expanding and solving for tan²θ:
9/4 - 1 = tan²θ
tan²θ = 9/4 - 4/4
tan²θ = 5/4

Taking the square root of both sides:
tanθ = ± √(5/4)
Since we know that tanθ < 0 in the third quadrant:
tanθ = - √(5/4)
tanθ = - √5/2

Finally, to find cotθ, we use the Pythagorean identity:

cot²θ + 1 = csc²θ

Substituting the values:
cot²θ + 1 = (-2/√5)²

Expanding:
cot²θ + 1 = 4/5

Solving for cot²θ:
cot²θ = 4/5 - 1
cot²θ = 4/5 - 5/5
cot²θ = -1/5

Taking the square root of both sides and remembering that cotθ < 0 in the third quadrant:
cotθ = - √(1/5)
cotθ = - 1/√5
Rationalizing the denominator:
cotθ = - √5/5

Therefore, the solutions are:
tanθ = - √5/2 and cotθ = - √5/5.

I hope this explanation helps!