We will learn step-by-step the proof of cotangent formula cot (α + β).
Prove that, cot (α + β) = cotαcotβ−1cotβ−cotα.
Proof: cot (α + β) = cos(α+β)sin(α+β)
= cosαcosβ−sinαsinβsinαcosβ+cosαsinβ
= cosαcosβsinαsinβ−sinαsinβsinαsinβsinαcosβsinαsinβ+cosαsinβsinαsinβ, [dividing numerator and denominator by sin α sin β].
= cotαcotβ−1cotβ−cotα. Proved
Therefore, cot (α + β) = cotαcotβ−1cotβ−cotα.
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 = cotxcot2x−1cot2x+cotx
⇒ 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 cotα(1+cotα) ∙ cotβ(1+cotβ) = 1/2
Solution:
Given, α + β = 225°
α + β = 180° + 45°
cot (α + β) = cot (180° + 45°), [taking cot on both the sides]
⇒ cotαcotβ−1cotα+cotβ = cot 45°
⇒ cotαcotβ−1cotα+cotβ = 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)
⇒ cotα(1+cotα) ∙ cotβ(1+cotβ) = 1/2 Proved
11 and 12 Grade Math
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