Question

My original math question was Tanx - Cotx = 2secx.
On the math sheet I am trying to complete, the final answer was given but not how to reach that answer. The answer I should reach is 201 degrees in the third Quadrant, and 339 in the forth question, making the related angle 21 degrees.

The problem I am having is finding the 21 degrees. Here are my steps below:

Tanxcotx =2secx

sinx/cosx - cosx/sinx = 2(1/cosx)

Sinxcosx/cosx - cos^2x/sinx = 2

sinx - cos^2x/sinx = 2

Sin^2x/sinx - cos^2x/sinx = 2(sinx/sinx)

Sin^2x/sinx - cos^2x/sinx - 2(sinx/sinx) = 0

Answer 1

Then I turn the Cos^2x into 1 - Sin^x

Now at:

(sin^2x - (1-sin^2x) - 2sinx)/sinx = 0

Then I rearrange the top so it's:

2sin^2x - 2sinx - 1 = 0

Now I use the Quadratic formula since I cannot factor anything at this point.

a = 2
b = -2
c = -1

After putting in my answers, I get the two results of:

0.3660 and -1.3660

Then I have to do the sinx inverse of both of then. With 0.03660, I get 21 degrees. But for me to get my answer in the third and forth quadrant, the 21 degrees needs to be negative.

-1.3660 does not work, it gives an error.

So something must be wrong and I'm not sure what could be wrong here. Anyone able to help?

Answer 2

\[\tan(x)*\cot(x) = 1\]
in your steps you turn it into tan - cot ?

Answer 3

Ahh my mistake the original question is Tanx - Cotx = 2secx. I forgot the subtraction sign.

Answer 4

your work is correct...21 degrees is the reference angle but if you plug it into original it doesn't work due to opposite signs
tan - cot is neg in quad 1 and 3 ................. 2sec is neg in quad 2 and 3
tan - cot is pos in quad 2 and 4 ................. 2 sec is pos in quad 1 and 4

thus solutions are in quad 3 and 4 using reference angle of 21 degrees

Answer 5

Ahh I see thank you.

Answer 6

yw