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






11 and 12 Grade Math

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