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Proof of Tangent Formula tan (α - β)

We will learn step-by-step the proof of tangent formula tan (α - β).

Prove that: tan (α - β) = tanαtanβ1+tanαtanβ.

Proof: tan (α - β) = sin(αβ)cos(αβ)

sinαcosβcosαsinβcosαcosβ+sinαsinβ

sinαcosβcosαcosβcosαsinβcosαcosβcosαcosBcosαcosβ+sinαsinβcosαcosβ, [dividing numerator and denominator by cos α cos β].

tanαtanβ1+tanαtanβ          Proved

Therefore, tan (α - β) = tanαtanβ1+tanαtanβ.

Solved examples using the proof of tangent formula tan (α - β):

1. Find the values of tan 15°

Solution:

tan 15° = tan (45° - 30°)

           = tan45°tan30°1+tan45°tan30°

           = 1131+(113)

           = 313+1

           = (31)2(3+1)(31)

           = (3)223+(1)2(3+1)(31)

           = 3+12331

           = 4232

           = 2 - √3

 

2. Prove the identities: cos10°sin10°cos10°+sin10° = tan 35°      

Solution:

L.H.S = cos10°sin10°cos10°+sin10°

        = 1tan10°1+tan10°, (dividing numerator and denominator by cos 10°)

        = tan45°tan10°1+tan45°tan10°, (Since we know that, tan 45° = 1)

        = tan (45° - 10°)

        = tan 35°              Proved

 

3. If x - y = π/4, prove that (1 + tan x)(1 + tan y) = 2 tan x

Solution:

Given, x - y = π/4

⇒ tan (x - y) = tan π/4

tanxtany1+tanxtany = 1, [since tan π/4 = 1]

⇒ 1 + tan x tan y = tan x - tan y

⇒ 1 + tan x tan y + tan y = tan x

⇒ 1 + tan x + tan x tan y + tan y = tan x + tan x, [Adding tan x to both the sides]

⇒ (1 + tan x)(1 + tan y) = 2 tan x              Proved

 

6. If tan β = nsinαcosα1nsin2α, show that tan (α - β) = (1 - n) tan α

Solution:

tan (α - β) = tanαtanβ1+tanαtanβ

= sinαcosαnsinαcosα1nsin2α1+sinαcosαnsinαcosα1nsin2α

sinα(1nsin2α)nsinαcos2αcosα(1nsin2α)+nsin2αcosα

= sinαcosα1nsin2αncos2α1nsin2α+nsin2α

= sinαcosα1(nsin2α+cos2α)1

= tan α ∙ (1 - n ∙ 1), [since, we know that sin2 θ + cos2 θ = 1]

= (1 - n) tan α              Proved

 

 7. If tan β = sinαcosα2+cos2α prove that 3 tan (α - β) = 2 tan α.

Solution:

We have, tan (α - β) = tanαtanβ1+tanαtanβ

⇒ tan (α - β) = sinαcosαsinαcosα2+cos2α1+sinαcosαsinαcosα2+cos2α, [since we know that, tan β = sinαcosα2+cos2α

⇒ tan (α - β) = 2sinα+sinαcos2αsinαcos2α2cosα+cos3α+sin2αcosα

 ⇒ tan (α - β) = 2sinαcosα(2+cos2α+sin2α)

⇒ tan (α - β) = 2sinαcosα(2+1), [since we know that cos2 θ + sin2 θ = 1]

⇒ tan (α - β) = 2sinα3cosα

⇒ tan (α - β) = 3 tan (α - β)

⇒ tan (α - β) = 2 tan α              Proved

 Compound Angle






11 and 12 Grade Math

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