Quadrant III, csc(theta)= -1.25, find tan(theta)
Quadrant IV, sec(theta)= 4/3, find cot(theta)
Please provide the work for both. Thank you!
I will do the first one, you do the second.
csc(theta) = -1.25 = -5/4 and we are in III
so sin(theta) = -4/5
construct your right-angles triangle, and since sin(theta) = y/r
y = -4, r = 5 and x = ???
x^2 + (-4)^2 = 5^2
x = -3 since we are in quadrant III
tan(theta) = -3/5
For the second problem, I got 4squareroot7/7
not quite -- 4/√7 is cscθ
did you draw the triangle?
To find tan(theta) when csc(theta) = -1.25 in Quadrant III, we can use the following steps:
1. Recall the trigonometric identity: csc(theta) = 1/sin(theta).
2. Therefore, sin(theta) = 1/csc(theta) = 1/(-1.25) = -0.8.
3. In Quadrant III, the sine function is negative.
4. Using the Pythagorean identity, cos^2(theta) = 1 - sin^2(theta), we can find cos(theta).
cos^2(theta) = 1 - (-0.8)^2 = 1 - 0.64 = 0.36.
Taking the square root of both sides, cos(theta) = ±√(0.36) = ±0.6.
5. Since we are in Quadrant III, the cosine function is negative.
Therefore, cos(theta) = -0.6.
6. Finally, tan(theta) = sin(theta)/cos(theta) = (-0.8)/(-0.6) = 4/3.
Now let's find cot(theta) when sec(theta) = 4/3 in Quadrant IV:
1. Recall the trigonometric identity: sec(theta) = 1/cos(theta).
2. Therefore, cos(theta) = 1/sec(theta) = 1/(4/3) = 3/4.
3. In Quadrant IV, the cosine function is positive.
4. Using the Pythagorean identity, sin^2(theta) = 1 - cos^2(theta), we can find sin(theta).
sin^2(theta) = 1 - (3/4)^2 = 1 - 9/16 = 7/16.
Taking the square root of both sides, sin(theta) = ±√(7/16) = ±√(7/16) = ±√7/4.
5. Since we are in Quadrant IV, the sine function is negative.
Therefore, sin(theta) = -√7/4.
6. Finally, cot(theta) = 1/tan(theta) = cos(theta)/sin(theta) = (3/4)/(-√7/4) = -3√7/7.