Question

Prove that: `(tan θ)/(1 - cot θ) + (cot θ)/(1 - tan θ) = 1 + tan θ + cot θ`.

Answer 1

Proof:

1. LHS and express cot θ in terms of tan θ:

LHS = `(tan θ)/(1 - cot θ) + (cot θ)/(1 - tan θ)`

Since `cot θ = 1/(tan θ)`:

LHS = `(tan θ)/(1 - 1/tan θ) + (1/tan θ)/(1 - tan θ)`

2. Simplify the denominators:

LHS = `(tan θ)/((tan θ - 1)/(tan θ)) + 1/(tan θ(1 - tan θ))`

LHS = `(tan^2 θ)/(tan θ - 1) + 1/(tan θ(1 - tan θ))`

3. Make the denominators common (Note that tan θ – 1 = – (1 – tan θ):

LHS = `(tan^2 θ)/(tan θ - 1) - 1/(tan θ(tan θ - 1))`

LHS = `(tan^3 θ - 1)/(tan θ(tan θ - 1))`

4. Apply the difference of cubes formula (a³ – b³ = (a – b)(a² + ab + b²)):

tan³ θ – 1 = (tan θ – 1)(tan² θ + tan θ + 1)

Substitute this back into the expression:

LHS = `((tan θ - 1)(tan^2 θ + tan θ + 1))/(tan θ(tan θ - 1))`

5. Cancel (tan θ – 1) and divide:

LHS = `(tan^2 θ + tan θ + 1)/(tan θ)`

LHS = `(tan^2 θ)/(tan θ) + (tan θ)/(tan θ) + 1/(tan θ)`

LHS = tan θ + 1 + cot θ

LHS = 1 + tan θ + cot θ = RHS.