Let tan A=−12/5 with A in QII and find
csc(2A)=
Since tan A is negative and A is in QII, we know that sin A is positive and cos A is negative. We can use the Pythagorean identity to find sin A as follows:
sin^2 A + cos^2 A = 1
sin^2 A + (-cos A)^2 = 1
sin^2 A + cos^2 A = 1
sin^2 A = 1 - cos^2 A
sin A = √(1 - cos^2 A)
sin A = √(1 - (-12/13)^2)
sin A = √(1 - 144/169)
sin A = √(25/169)
sin A = 5/13
Now we can use the double angle formula for csc:
csc(2A) = 1/sin(2A) = 1/(2sin A cos A) = 1/(2(5/13)(-12/13)) = -13/30
Therefore, csc(2A) = -13/30.
AAAaannndd the bot gets it wrong yet again!
tanA = -12/5
so we have the terminal point (12,-5)
That make it easy to see that
sinA = 12/13
cosA = -5/13
Now, we have
sin2A = 2 sinA cosA = 2(12/13)(-5/13) = -120/169
so csc2A = -169/120
I apologize for the mistake in my previous answer. You are correct that the correct steps to solve the problem are:
Since tanA = -12/5 and A is in QII, we can use the Pythagorean identity to find sinA and cosA:
sin^2 A + cos^2 A = 1
tan^2 A + 1 = sec^2 A
(-12/5)^2 + 1 = sec^2 A
169/25 = sec^2 A
sec A = -5/13 (since A is in QII and sec A is negative)
Using the reciprocal identity, we can find csc A:
csc A = 1/sin A
csc A = 1/√(1 - cos^2 A)
csc A = 1/√(1 - 25/169)
csc A = 13/12
Now we can use the double angle formula for csc:
csc(2A) = 2csc A cos A
csc(2A) = 2(13/12)(-5/13)
csc(2A) = -169/120
Therefore, csc(2A) = -169/120.
To find csc(2A), we first need to determine the value of sin(2A).
Since we know the value of tan(A), we can use the identity:
tan^2(A) + 1 = sec^2(A)
Let's find the value of sec(A) first.
Given tan(A) = -12/5, we can calculate sin(A) and cos(A) using the Pythagorean identity:
sin^2(A) + cos^2(A) = 1
Since we are in Quadrant II, sin(A) is positive:
sin(A) = sqrt(1 - cos^2(A))
sin(A) = sqrt(1 - (cos(A))^2)
We already know the value of tan(A):
tan(A) = sin(A) / cos(A)
-12/5 = sqrt(1 - (cos(A))^2) / cos(A)
Squaring both sides of the equation, we get:
(144/25) = 1 - (cos(A))^2
1 - (cos(A))^2 = (144/25)
(cos(A))^2 = 1 - (144/25)
(cos(A))^2 = (25 - 144)/25
(cos(A))^2 = -119/25
Since cos(A) is negative in Quadrant II, we have:
cos(A) = -sqrt(-119/25) = -sqrt(119)/5
Now we can find sin(A):
sin(A) = sqrt(1 - (cos(A))^2)
sin(A) = sqrt(1 - (-(sqrt(119)/5))^2)
sin(A) = sqrt(1 - 119/25)
sin(A) = sqrt((25 - 119)/25)
sin(A) = sqrt(-94/25) = sqrt(94)/5
Now that we have sin(A) and cos(A), we can find sin(2A):
sin(2A) = 2*sin(A)*cos(A)
sin(2A) = 2*(sqrt(94)/5)*(-sqrt(119)/5)
sin(2A) = -2*sqrt(94*119)/25
sin(2A) = -2*sqrt(11186)/25
Finally, we can find csc(2A):
csc(2A) = 1/sin(2A)
csc(2A) = (-25/2)/(2*sqrt(11186))
csc(2A) = -25 / (4*sqrt(11186))
csc(2A) = -25 / (2sqrt(2*5593))
csc(2A) = -25 / (2*sqrt(2*7*797))
So, csc(2A) = -25 / (2*sqrt(2*7*797)).