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We will learn step-by-step the proof of tangent formula tan (α - β).
Prove that: tan (α - β) = tanα−tanβ1+tanαtanβ.
Proof: tan (α - β) = sin(α−β)cos(α−β)
= sinαcosβ−cosαsinβcosαcosβ+sinαsinβ
= sinαcosβcosαcosβ−cosαsinβcosαcosβcosαcosBcosαcosβ+sinαsinβcosαcosβ, [dividing numerator and denominator by cos α cos β].
= tanα−tanβ1+tanαtanβ Proved
Therefore, tan (α - β) = tanα−tanβ1+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°
= 1−1√31+(1∙1√3)
= √3−1√3+1
= (√3−1)2(√3+1)(√3−1)
= (√3)2−2∙√3+(1)2(√3+1)(√3−1)
= 3+1−2∙√33−1
= 4−2√32
= 2 - √3
2. Prove the identities: cos10°−sin10°cos10°+sin10° = tan 35°
Solution:
L.H.S = cos10°−sin10°cos10°+sin10°
= 1−tan10°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
⇒ tanx−tany1+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 = nsinαcosα1−nsin2α, show that tan (α - β) = (1 - n) tan α
Solution:
tan (α - β) = tanα−tanβ1+tanαtanβ
= sinαcosα−nsinαcosα1−nsin2α1+sinαcosα⋅nsinαcosα1−nsin2α
= sinα(1−nsin2α)−nsinαcos2αcosα(1−nsin2α)+nsin2αcosα
= sinαcosα⋅1−nsin2α−ncos2α1−nsin2α+nsin2α
= sinαcosα⋅1−(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) = tanα–tanβ1+tanαtanβ
⇒ tan (α - β) = sinαcosα−sinαcosα2+cos2α1+sinαcosα∙sinαcosα2+cos2α, [since we know that, tan β = sinαcosα2+cos2α
⇒ tan (α - β) = 2sinα+sinαcos2α−sinαcos2α2cosα+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
11 and 12 Grade Math
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