Let tan A=−12/5 with A in QII and find

csc(2A)=

3 answers

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.