Question

Use the Pythagorean indentities rather than reference triangles. Find tan θ and cot θ if sec θ=8/7 and sin θ <0

Answer 1

sec θ = 1/cos θ = 8/7
sin is negative and cos is positive so in quadrant IV, lower right below +x axis

Identity 3 ---> sec^2 = 1 + tan^2
so
64/49 = 1 + tan^2
tan^2 = 64/49 - 49/49 = 15/49
so knowing it is negative
tan θ = -(1/7)sqrt 15

now use identity 2
1 + cot^2 = csc^2
to get cot
no fair saying cot = 1 / tan :)

Answer 2

1/csc^2 + 1/sec^2 = 1
sec^2 + csc^2 = csc^2 sec^2
64/49 + csc^2 = csc^2 (64/49)
(64 -49)/49 csc^2 = 64/49
15 csc^2 = 64
csc^2 = 64/15

cot^2 = csc^2 - 1 = (64-15)/15
= 49/15
cot = -7/sqrt15 negative because in quad IV= - (7/15)sqrt 15
but we knew that :)

Answer 3

sec(A) = 8/7 = 1/Cos(A).
1/Cos(A) = 8/7,
Cos(A) = 7/8 = X/r,
X = 7, Y = ?, r = 8.

x^2 + y^2 = r^2.
7^2 + y^2 = 8^2,
y^2 = 8^2-7^2 = 15,
Y = sqrt(15).
sin(A)<0(neg.), Then Y = -sqrt(15).

Tan(A) = Y/X = -(sqrt(15)/7)-.

Cot(A) = 1/Tan(A) = 7/-sqrt(15),
Multiply numerator and denominator by sqrt(15):
Cot(A) =-(7*sqrt(15)/15).










































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