Question

How do I solve this?
tan(x- 10) = cot(2x + 17)

Answer 1

Rewrite cot(A) as tan(pi/2 - A).

Answer 2

How does that work?

Answer 3

tan(x- 10) = cot(2x + 17)
cot(2x+17) = tan(pi/2 - 2x - 17)
tan(x- 10) = tan(pi/2 - 2x - 17)
x - 10 = pi/2 - 2x - 17
solve for x.
And since tan has a period of pi, the general solution will be whatever you get for x above plus/minus pi.

Answer 4

plus/minus N*pi.

Answer 5

So since they're reciprocals we can just subtract by 180

Answer 6

All "co" trig functions are pi/2 or 90 degree phase sifted from the corresponding trig function.

sin(A) = "co"sine(pi/2 - A)
"co"sine(A) = sin(pi/2 - A)

sec(A) ="co"sec(pi/2-A)
"co"sec(A) = sec(pi/2-A)

tan(A) = "co"tan(pi/2-A)
"co"tan(A) = tan(pi/2-A)

Answer 7

OR,
sin(A) = cos(pi/2-A)
cos(A) = sin(pi/2-A)
sec(A) = cosec(pi/2-A)
cosec(A) = sec(pi/2-A)
tan(A) = cot(pi/2-A)
cot(A) = tan(pi/2-A)

Answer 8

*phase shifted*

Answer 9

Also, you will get a second set of solutions by another identity:
cot(A) = -tan(pi/2 + A) = tan(-pi/2 - A)

Answer 10

Thank you sooo much

Answer 11

You are welcome.