Arun
Last Activity: 6 Years ago
There are two cases for this problem
Case 1 : x > 0
Case 2 : x
Case 1:
x > 0
let theta = tan^-1(1/x) where theta = (0, pi/2) i.e. not including 0 and pi/2
Then
tan(theta) = 1/x
cot(theta) = x
When x > 0,
theta = cot^-1(x)
Case 2:
x
Let theta = tan^-1(1/x) where theta = (pi/2, pi)
Then
tan(theta) = 1/x
-tan(pi - theta) = 1/x
tan(theta - pi) = 1/x
cot(theta - pi) = x
theta - pi = cot^-1(-x)
theta = -pi + cot^-1(-x)
Therefore,
tan^-1(1/x) = -pi + cot^-1(-x)