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
11 and 12 Grade Math
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