# Prove (show complete solution and explain the answer): Tanθ/(1-Cotθ) + Cotθ/(1-Tanθ) = 1 + Tanθ + Cotθ

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You may process either one of the sides, or both sides together. I will process the left side.

You need to bring the fractions to the left side to a common denominator such that:

`((tan theta)(1 - tan theta) + (cot theta)(1 - cot theta))/((1 - tan theta)(1 - cot theta)) = 1 + tan theta + cot theta`

Opening the brackets yields:

`(tan theta - tan^2 theta + cot theta - cot^2 theta)/((1 - tan theta)(1 - cot theta)) = 1 + tan theta + cot theta`

You need to use the fact that cot theta= 1/tan theta such that:

`(tan theta - tan^2 theta +1/tan theta- 1/(tan^2 theta))/((1 - tan theta)(1 - 1/tan theta)) = 1 + tan theta + 1/tan theta`

`((tan^3 theta - tan^4 theta + tan theta - 1)/(tan theta))/(((1 - tan theta)(tan theta- 1))/(tan theta)) = 1 + tan theta + 1/tan theta`

Reducing by tan theta yields:

`(tan^3 theta - tan^4 theta + tan theta - 1)/(tan theta(1 - tan theta)(tan theta- 1)) = 1 + tan theta + 1/tan theta`

Factoring out `tan^3 theta` yields:

`(tan^3 theta(1 - tan theta)- (1 - tan theta))/(tan theta(1 - tan theta)(tan theta- 1)) = 1 + tan theta + 1/tan theta`

Factoring out `(1 - tan theta)` yields:

`((1 - tan theta)(tan^3 theta - 1))/(tan theta(1 - tan theta)(tan theta- 1)) =1 + tan theta + 1/tan theta`

Using the special product `(tan theta- 1)(tan^2 theta + tan theta + 1)` instead of difference of cubes `tan^3 theta - 1` yields:

`((1 - tan theta)(tan theta- 1)(tan^2 theta + tan theta + 1))/(tan theta(1 - tan theta)(tan theta- 1)) =1 + tan theta + 1/tan theta`

Reducing by `(1 - tan theta)(tan theta- 1) ` yields:

`(tan^2 theta + tan theta + 1)/tan theta= 1 + tan theta + 1/tan theta`

`(tan^2 theta)/tan theta + tan theta/tan theta + 1/tan theta = 1 + tan theta + 1/tan theta`

Reducing by `tan theta ` to the left side yields:

`tan theta + 1 + 1/tan theta = 1 + tan theta + 1/tan theta`

`` Notice that most of transformations were done to the left side. The right side suffers only one change, such that `cot theta = 1/tan theta.`

**Hence, the last line proves the identity `tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + tan theta + cot theta.` **