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) = $$\frac{tan α - tan β}{1 + tan α tan β}$$.

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

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