TRIGONOMETRY QUESTION AND ANSWER FOR GRADE 11

Question 1 :

Prove that tan (π/4 + θ) tan (3π/4 + θ)  =  −1.

Answer :

tan (π/4 + θ)  =  (tan π/4 + tan θ) / (1 - tan π/4 tan θ)

tan (π/4 + θ)  =  (1 + tan θ) / (1 - tan θ)  -----(1)

tan (3π/4 + θ)  =  (tan 3π/4 + tan θ) / (1 - tan 3π/4 tan θ)

=  (-1 + tan θ) / (1 - (-1) tan θ)

tan (3π/4 + θ)  =  (tan θ - 1) / (1 + tan θ)   -----(2)

(1) ⋅ (2)

=  [(1 + tan θ) / (1 - tan θ)]  ⋅ [(tan θ - 1) / (1 + tan θ) ]

=   (tan θ - 1) /  (1 - tan θ)

=  -1

Hence proved.

Question 2 :

Find the values of tan(α+β), given that cot α = 1/2, α ∈ (π, 3π/2) and sec β = −5/3, β ∈  (π/2, π).

Answer : 

tan (α+β)  =  (tan α + tan β) / (1 - tan α tan β)

cot α = 1/2

α lies in 3rd quadrant

tan α  =  2

sec β = −5/3

β lies in 2nd quadrant

tan β  =  √(sec2 β - 1)

  =  √(-5/3)2- 1)

  =  √(25/9)- 1)

  =  √16/9

tan β  = - 4/3

  =  (2 + (-4/3)) / (1 - 2 (-4/3))

  =  (6 - 4)/3 / (1 + (8/3))

  =  (2/3)  / (11/3)

  =  2/11

Hence the answer is 2/11.

Question 3 :

 If θ + φ = α and tan θ = k tan φ, then prove that

sin(θ − φ)  =  [(k − 1)/(k + 1)]sin α

Answer :

Given that :

θ + φ  =  α and tan θ  =  k tan φ

tan θ/tan φ  =  k /1

a/b  =  c/d  ==>  (a - b)/(a + b)  =  (c - d)/(c + d)

tan θ/tan φ  =  k /1

(tan θ - tan φ) / (tan θ + tan φ)  =  (k - 1)/(k + 1)  ---(1)

tan θ - tan φ  =  (sin θ / cos θ) - (sin φ / cos φ)

  =  (sin θcos φ - cos θsin φ) / cos θ cos φ

tan θ - tan φ  =  sin (θ - φ)

tan θ + tan φ  =  (sin θ / cos θ) + (sin φ / cos φ)

  =  (sin θcos φ + cos θsin φ) / cos θ cos φ

tan θ + tan φ  =  sin (θ + φ)

tan θ + tan φ  =  sin α

Applying the values of tan θ - tan φ and tan θ + tan φ in (1), we get

sin (θ - φ)/sin α  =  (k - 1)/(k + 1)

sin (θ - φ)  =  [(k - 1)/(k + 1)] sin α

Hence proved

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