# Proof of Cotangent Formula cot (α + β)

We will learn step-by-step the proof of cotangent formula cot (α + β).

Prove that, cot (α + β) = $$\frac{cot α cot β - 1}{cot β - cot α}$$.

Proof: cot (α + β) = $$\frac{cos (α + β)}{sin (α + β)}$$

= $$\frac{cos α cos β - sin α sin β}{sin α cos β + cos α sin β}$$

= $$\frac{\frac{cos α cos β}{sin α sin β} - \frac{sin α sin β}{sin α sin β}}{\frac{sin α cos β}{sin α sin β} + \frac{cos α sin β}{sin α sin β}}$$, [dividing numerator and denominator by sin α sin β].

= $$\frac{cot α cot β - 1}{cot β - cot α}$$.            Proved

Therefore, cot (α + β) = $$\frac{cot α cot β - 1}{cot β - cot α}$$.

Solved examples using the proof of cotangent formula cot (α + β):

1. Prove the identities: cot x cot 2x - cot 2x cot 3x - cot 3x cot x = 1

Solution:

We know that 3x = 2x + x

Therefore, cot 3x = cot (x + 2x)

cot 3x = $$\frac{cot x cot 2x - 1}{cot 2x + cot x}$$

⇒ cot x cot 2x - 1 = cot 2x cot 3x + cot 3x cot x

⇒ cot x cot 2x - cot 2x cot 3x - cot 3x cot x = 1            Proved

2. If α + β = 225° show that $$\frac{cot α}{(1 + cot α)}$$ ∙ $$\frac{cot β}{(1 + cot β)}$$ = 1/2

Solution:

Given, α + β = 225°

α + β = 180° + 45°

cot (α + β) = cot (180° + 45°), [taking cot on both the sides]

⇒ $$\frac{cot α cot β - 1}{cot α + cot β}$$ = cot 45°

⇒ $$\frac{cot α cot β - 1}{cot α + cot β}$$ = 1, [since we know cot 45° = 1]

⇒ cot α cot β - 1 = cot α + cot β

⇒ cot α cot β = 1 + cot α + cot β

⇒ 2 cot α cot β = 1 + cot α + cot β + cot α cot β, [adding cot α cot β on both sides]

⇒ 2 cot α cot β = (1 + cot α) + cot β (1 + cot α)

⇒ 2 cot α cot β = (1 + cot α) + cot β (1 + cot α)

⇒ 2 cot α cot β = (1 + cot α)(1 + cot β)

⇒ $$\frac{cot α}{(1 + cot α)}$$ ∙ $$\frac{cot β}{(1 + cot β)}$$ = 1/2            Proved