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Obtuse and reflex angles.

Hey guys,

I'm stuck on a question.
Given that secA=3 and the angle A is reflex, find the exact value of:
a) tanA
b) cotA
c) cosecA


a) For this part I used the triangle method and got 2sqrt2. Then looking at the CAST diagram, tan is positive in the bottom left (4th?) quadrant. So the answer should be positive right? My answer booklet say it is negative.

b) I'm assuming that the signs are the same for inverse functions as they are for the regular ones, is that correct? So the answer should be 1/(2sqrt2) or (sqrt2)/4. Again the answer booklet says it's negative, I can't understand why.

c) This is the only answer I agree on as sinA is negative in this quadrant so (again assuming signs are the same for the inverse functions) cosecA must be negative. I got -(3sqrt2)/4.

If someone could explain where I'm going wrong I would be grateful, thanks!
Original post by Supertod
Hey guys,

I'm stuck on a question.


I'll deal with them all at once.

"I'm assuming that the signs are the same for inverse functions as they are for the regular ones"

Yes, Cos & sec will have the same sign since sec = 1/cos. Similarly cot & tan, and sin & cosec.

Now your angle is reflex, so it lies in quadrants 3 or 4, lower left or lower right.

sec is positive, so cos is positive, and so your angle, A, lies in the lower right, 4th quadrant.
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Supertod
OP
Original post by ghostwalker
I'll deal with them all at once.

"I'm assuming that the signs are the same for inverse functions as they are for the regular ones"

Yes, Cos & sec will have the same sign since sec = 1/cos. Similarly cot & tan, and sin & cosec.

Now your angle is reflex, so it lies in quadrants 3 or 4, lower left or lower right.

sec is positive, so cos is positive, and so your angle, A, lies in the lower right, 4th quadrant.


Ahh I get you, I didn't realise reflex could be either. Thanks!

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