Use the Pythagorean indentities rather than reference triangles. Find tan θ and cot θ if sec θ=8/7 and sin θ <0
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 :)
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 :)
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).
0
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To find tan θ and cot θ using the Pythagorean identities, we need to begin by finding the value of sin θ.
Given that sec θ = 8/7, we can use the formula: sec θ = 1/cos θ.
So, 1/cos θ = 8/7.
To solve for cos θ, we can cross-multiply:
7 = 8cos θ.
Divide both sides by 8:
cos θ = 7/8.
Now, we can use the Pythagorean identity to find sin θ:
sin^2 θ + cos^2 θ = 1.
Substituting the value of cos θ:
sin^2 θ + (7/8)^2 = 1.
sin^2 θ + 49/64 = 1.
To isolate sin^2 θ, we subtract 49/64 from both sides:
sin^2 θ = 1 - 49/64.
sin^2 θ = (64/64) - (49/64).
sin^2 θ = 15/64.
Since sin θ < 0, we take the negative square root:
sin θ = -√(15/64).
Now, we can find tan θ using the identity: tan θ = sin θ/cos θ.
tan θ = (-√15/8) / (7/8).
Dividing both numerator and denominator by 1/8:
tan θ = -√15 / 7.
Lastly, we can find cot θ using the identity: cot θ = 1/tan θ.
cot θ = 1 / (-√15 / 7).
To divide by a fraction, we multiply by its reciprocal:
cot θ = 7 / (-√15).
Therefore, tan θ = -√15 / 7 and cot θ = -7 / √15.