How do you prove #cot^-1(x)=pi/2-tan^-1(x)#?

1 Answer

Kindly see the solution for the answer

Explanation:

The solution

Let #alpha# be the angle with tangent#=x#
Let #beta=pi/2-alpha# be the angle with cotangent #=x#

Then #alpha+beta=pi/2#

#cot^-1 x=pi/2-tan^-1 x#

#cot^-1 x=pi/2-tan^-1 x#

#cot^-1 x=pi/2-(pi/2-cot^-1 x)#

#cot^-1 x=pi/2-pi/2+cot^-1 x#

#cot^-1 x=cot^-1 x#

God bless....I hope the explanation is useful.