sec(t) = -2.6 = -26/10 = -13/5
so cos(t) = -5/13
construct your right-angled triangle with hypotenuse 13 and adjacent side of 5
( remember cosØ = adjacent/hypotenuse )
so 5^2 + y^2 = 13^2
y^2 = 169-25 = 144
y = ± 12
but for π < t < 3π/2 , y = -12
Tan(t) = opposite/ adjacent = -12/5
Let t be an angle between π and 3π/2, and suppose sec(t)=−2.6. What is tan(t)?
Having a hard time with this one..
6 answers
thank you so much!!! Also, I have one more question..
Assume r>0. Which of the following equals csc(tan−1(r))?
Is sqrt1+r2 the correct answer
Assume r>0. Which of the following equals csc(tan−1(r))?
Is sqrt1+r2 the correct answer
angle t is in quadrant II, so the x is negative, and the y is positive,
so x = -5, and y = 12
does not change the final answer of tan(t) = -12/5
so x = -5, and y = 12
does not change the final answer of tan(t) = -12/5
csc(tan−1(r)) = ???
start with tan^-1 (r) = tan^-1 (r/1)
so you have a triangle with opposite of r and adjacent of 1
then hypotenuse^2 = r^2 + 1, hyp = √(r^2+1)
so sin(tan^−1 (r)) = opposite/hypo = r/√(r^2+1)
and csc(tan^−1 (r)) = √(r^2+1) / r
start with tan^-1 (r) = tan^-1 (r/1)
so you have a triangle with opposite of r and adjacent of 1
then hypotenuse^2 = r^2 + 1, hyp = √(r^2+1)
so sin(tan^−1 (r)) = opposite/hypo = r/√(r^2+1)
and csc(tan^−1 (r)) = √(r^2+1) / r
Quad III ... tan is positive
identity ... tan^2 = sec^2 - 1
identity ... tan^2 = sec^2 - 1
scott is correct, the angle was in III , (time to go to bed)
make the necessary changes, or follow scott's relation
tan^2 = sec^2 - 1
= (-13/5)^2 - 1 = 144/25
= + 12/5 in quad III
make the necessary changes, or follow scott's relation
tan^2 = sec^2 - 1
= (-13/5)^2 - 1 = 144/25
= + 12/5 in quad III