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

 Compound Angle






11 and 12 Grade Math

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