We will learn step-by-step the proof of tangent formula tan (α - β).
Prove that: tan (α - β) = \(\frac{tan α - tan β}{1 + tan α tan β}\).
Proof: tan (α - β) = \(\frac{sin (α - β)}{cos (α - β)}\)
= \(\frac{sin α cos β - cos α sin β}{cos α cos β + sin α sin β}\)
= \(\frac{\frac{sin α cos β}{cos α cos β} - \frac{cos α sin β}{cos α cos β}}{\frac{cos α cos B}{cos α cos β} + \frac{sin α sin β}{cos α cos β}}\), [dividing numerator and denominator by cos α cos β].
= \(\frac{tan α - tan β}{1 + tan α tan β}\) Proved
Therefore, tan (α - β) = \(\frac{tan α - tan β}{1 + tan α tan β}\).
Solved
examples using the proof of
tangent formula tan (α -
β):
1. Find the values of tan 15°
Solution:
tan 15° = tan (45° - 30°)
= \(\frac{tan 45° - tan 30°}{1 + tan 45° tan 30° }\)
= \(\frac{1 - \frac{1}{√3}}{1 + (1 ∙ \frac{1}{√3})}\)
= \(\frac{√3 - 1}{√3 + 1}\)
= \(\frac{(√3 - 1)^{2}}{(√3 + 1)(√3 - 1)}\)
= \(\frac{(√3)^{2} - 2 ∙ √3 + (1)^{2}}{(√3 + 1)(√3 - 1)}\)
= \(\frac{3 + 1 - 2 ∙ √3}{3 - 1}\)
= \(\frac{4 - 2√3}{2}\)
= 2 - √3
2. Prove the identities: \(\frac{cos 10° - sin 10°}{cos 10° + sin 10°}\) = tan 35°
Solution:
L.H.S = \(\frac{cos 10° - sin 10°}{cos 10° + sin 10°}\)
= \(\frac{1 - tan 10°}{1 + tan 10°}\), (dividing numerator and denominator by cos 10°)
= \(\frac{tan 45° - tan 10°}{1 + tan 45° tan 10°}\), (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
⇒ \(\frac{tan x - tan y}{1 + tan x tan y}\) = 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 = \(\frac{n sin \alpha cos \alpha}{1 - n sin^{2} \alpha}\), show that tan (α - β) = (1 - n) tan α
Solution:
tan (α - β) = \(\frac{tan \alpha - tan \beta }{1 + tan \alpha tan \beta}\)
= \(\frac{\frac{sin \alpha }{cos \alpha} - \frac{n sin \alpha cos \alpha}{1 - n sin^{2} \alpha}}{1 + \frac{sin \alpha}{cos \alpha}\cdot \frac{n sin \alpha cos \alpha}{1 - n sin^{2} \alpha}}\)
= \(\frac{sin \alpha (1 - n sin^{2} \alpha) - n sin \alpha cos^{2} \alpha}{cos \alpha (1 - n sin^{2} \alpha) + n sin^{2} \alpha cos \alpha}\)
= \(\frac{sin \alpha}{cos \alpha} \cdot \frac{1 - n sin^{2} \alpha - n cos^{2} \alpha}{1 - n sin^{2} \alpha + n sin^{2} \alpha}\)
= \(\frac{sin \alpha}{cos \alpha} \cdot \frac{1 - (n sin^{2} \alpha + cos^{2} \alpha)}{1 }\)
= tan α ∙ (1 - n ∙ 1), [since, we know that sin\(^{2}\) θ + cos\(^{2}\) θ = 1]
= (1 - n) tan α Proved
7. If tan β = \(\frac{sin α cos α}{2 + cos^{2} α}\) prove that 3 tan (α - β) = 2 tan α.
Solution:
We have, tan (α - β) = \(\frac{tan α – tan β}{1 + tan α tan β}\)
⇒ tan (α - β) = \(\frac{\frac{sin α}{cos α} - \frac{sin α cos α}{2 + cos^{2} α}}{1 + \frac{sin α}{cos α} ∙ \frac{sin α cos α}{2 + cos^{2} α}}\), [since we know that, tan β = \(\frac{sin α cos α}{2 + cos^{2} α}\)
⇒ tan (α - β) = \(\frac{2 sin α + sin α cos^{2} α - sin α cos^{2} α}{2 cos α + cos^{3} α + sin^{2} α cos α}\)
⇒ tan (α - β) = \(\frac{2 sin α}{cos α (2 + cos^{2} α + sin^{2} α)}\)
⇒ tan (α - β) = \(\frac{2 sin α}{cos α (2 + 1) }\), [since we know that cos\(^{2}\) θ + sin\(^{2}\) θ = 1]
⇒ tan (α - β) = \(\frac{2 sin α}{3 cos α}\)
⇒ tan (α - β) = 3 tan (α - β)
⇒ tan (α - β) = 2 tan α Proved
11 and 12 Grade Math
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