tanx = cotx?

Question

0<=x>2pi

Answer 1

Rewrite as tanx = 1/tanx

Cross multiply to get tan^2 x = 1

So tan x = 1 or tanx = -1

That happens at odd multiples of pi/4:

pi/4, 3pi/4, 5pi/4, 7pi/4

Answer 2

I think you mean 0<=x<= 2pi
This occurs at (1/4)pi, (3/4)pi, (5/4)pi, (7/4)pi
In other words, it happens whenever sinx = 1. Because the multiplicative inverse of 1 is 1.

Answer 3

tan(x) = cot(x)

Step 1: convert everything to sin and cos

sin(x)/cos(x) = cos(x)/sin(x)

Now, multiply both sides by the lowest common denominator,
sin(x)cos(x)

sin^2(x) = cos^2(x)

Use the identity cos^2(x) = 1 - sin^2(x)

sin^2(x) = 1 - sin^2(x)

Move the sin^2(x) to the left hand side

2sin^2(x) = 1
sin^2(x) = 1/2
Whenever we take the square root of both sides, we have to put a "plus or minus" on the right hand side. So

sin(x) = +/- 1/sqrt(2)

So sin(x) = 1/sqrt(2) and sin(x) = -1/sqrt(2).

Where is sin(x) equal to 1/sqrt(2)? We actually know this; based on the unit circle, this is true at x = pi/4 and 3pi/4.
Where is sin(x) equal to -1/sqrt(2)? At x = 5pi/4 and x = 7pi/4

Therefore, our solution is

x = pi/4, 3pi/4, 5pi/4, 7pi/4

Answer 4

no

tanx= 1/cotx

they have the same range (I believe).

simularly-
cotx=1/tanx