PRACTICE PROBLEMS IN TRIGONOMETRY FOR GRADE 10
Problem 1 :
Determine whether the following is an identity or not.
cos2θ + sec2θ = 2 + sinθ
Solution :
cos2θ + sec2θ :
= 1 - sin2θ + 1 + tan2θ
= 2 + tan2θ - sin2θ
≠ 2 + sinθ
Because cos2θ + sec2θ ≠ 2 + sinθ, it's not an identity.
Problem 2 :
Determine whether the following is an identity or not.
cot2θ + cos θ = sin2θ
Solution :
cot2θ + cosθ :
= cosec2θ - 1 + cosθ
≠ sin2θ
Because cot2θ + cos θ ≠ sin2θ, it's not an identity.
Problem 3 :
Prove the following identity.
sec2θ + cosec2θ = sec2θ ⋅ cosec2θ
Solution :
sec2θ + cosec2θ :
= (1/cos2θ) + (1/sin2θ)
= (sin2θ + cos2θ)/(cos2θ ⋅ sin2θ)
Value of sin2θ + cos2θ = 1.
Then,
= 1/(cos2θ sin2θ)
= (1/cos2θ)(1/sin2θ)
= sec2θ ⋅ cosec2θ
Problem 4 :
Prove the following identity.
sin θ /(1 - cos θ) = cosec θ + cot θ
Solution :
sin θ /(1-cos θ) :
Multiply both numerator and denominator by the conjugate of the denominator.
= [sin θ /(1-cos θ)] ⋅ [(1+cos θ)/(1+cos θ)]
Instead of (1 + cos θ)(1 + cos θ) we can write 1 - cos2θ by using the algebraic formula.
= [sin θ(1 + cosθ)]/(1-cos2θ)
= [sin θ(1 + cosθ)]/sin2θ
= (1 + cosθ)]/sinθ
= (1/sinθ) + (cosθ/sinθ)
= cosec θ + cot θ
Problem 5 :
Prove the following identity.
√(1 - sin θ)/(1 + sin θ) = sec θ - tan θ
Solution :
√(1 - sin θ)/(1 + sin θ) :
= √(1 - sin θ)/(1 + sin θ) ⋅ (1-sin θ)/(1-sin θ)
= √(1 - sin θ)2/[(1 + sin θ) x (1 - sin θ)]
= √(1 - sin θ)2/(12- sin2θ)
= √(1 - sin θ)2/(cos2θ)
= √[(1 - sin θ)/(cosθ)]2
= [(1 - sin θ)/(cosθ)]
= [(1/cosθ) - (sin θ/cosθ)]
= sec θ - tan θ
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