Example 1 :
Prove the following identities.
(i) sec4θ (1 - sin4θ) - 2tan2θ = 1
Solution :
sec4θ (1 - sin4θ) - 2tan2θ :
= sec4θ [(12)2 - (sin2θ)2] - 2tan2θ
= sec4θ [(1 + sin2θ) (1 - sin2θ)] - 2tan2θ
= (1/cos4θ)[(1 + sin2θ) cos2θ] - 2tan2θ
= [(1 + sin2θ)/cos2θ] - 2tan2θ
= (1/cos2θ) + (sin2θ/cos2θ) - 2tan2θ
= sec2θ + tan2θ - 2tan2θ
= sec2θ - tan2θ
= 1
Hence proved.
(ii) (cot θ-cos θ)/(cot θ+cos θ) = (cosec θ-1)/(cosec θ+1)
Solution :
(cot θ - cos θ)/(cot θ + cos θ) :
= ((cos θ/sin θ) - cos θ) / ((cos θ/sin θ) + cos θ)
= ((cos θ-sin θ cos θ)/sin θ) / (cos θ + sin θ cos θ)/sin θ)
= ((cos θ-sin θ cos θ) / (cos θ + sin θ cos θ)
= cos θ(1 - sin θ) / cos θ(1 + sin θ)
= (1 - sin θ)/(1 + sin θ)
= [1 - (1/cosec θ)] / [1 + (1/cosec θ)]
= (cosec θ - 1) / (cosec θ + 1)
Hence proved.
Example 2 :
Prove the following identities.
(i) [(sin A - sin B)/(cos A + cos B)] + [(cos A - cos B)/sin A + sin B)] = 0
Solution :
= (1 - 1)/(cos A + cos B) (sin A + sin B)
= 0
Hence proved.
(ii) [(sin3A + cos3A)/(sin A + cos A)] + [(sin3A - cos3A)/(sin A - cos A)] = 2
Solution :
[(sin3A + cos3A)/(sin A + cos A)] + [(sin3A - cos3A)/(sin A - cos A)]
a3 + b3 = (a + b)(a2 - ab + b2)
a3 - b3 = (a - b)(a2 + ab + b2)
[(sin3A + cos3A)/(sin A + cos A)]
= (sin A+cos A)(sin2A+cos2A-sinAcosA)/(sinA+cosA)
= (sin2A + cos2A - sin A cos A)
= (1 - sin A cos A) -------(1)
[(sin3A - cos3A)/(sin A - cos A)]
= (sin A-cos A)(sin2A+cos2A+sinAcosA)/(sinA-cosA)
= (sin2A + cos2A + sin A cos A)
= (1 + sin A cos A) -------(2)
(1) + (2)
= 1 - sin A cos A + 1 + sin A cos A
= 2
Hence proved.
Example 3 :
(i) If sin θ + cos θ = √3 , then prove that
tan θ + cot θ = 1
Solution :
Given : sin θ + cos θ = √3.
(sin θ + cos θ)2 = 3
sin2θ + cos2θ + 2sin θ cos θ = 3
2sin θ cos θ = 3 - 1
sin θ cos θ = 1
tan θ + cot θ :
= (sin θ/cos θ) + (cos θ/sin θ)
= (sin2 θ + cos2θ)/(sin θ cos θ)
= 1/(sin θ cos θ)
= 1/1
= 1 R.H.S
Hence proved.
(ii) If √3 sin θ − cos θ = 0, then show that
tan 3θ = (3 tan θ - tan3θ)/(1 - 3tan2θ)
Solution :
Given : √3 sin θ − cos θ = 0.
Then,
√3 sin θ = cos θ
sin θ/cos θ = 1/√3
tan θ = 1/√3
θ = 30°
tan 3θ :
= tan 3(30°)
= tan 90°
= undefined -----(1)
(3 tan θ - tan3θ)/(1 - 3tan2θ) :
= (3tan 30° - tan330°) / (1 - 3tan230°)
= [3(1/√3) - (1/√3)3] / [1 - 3(1/√3)2)]
= [√3 - (1/3√3)] / [1 - 3(1/3)]
= [√3 - (1/3√3)] / [1 - 1]
= [√3 - (1/3√3)] / 0
= undefined -----(2)
From (1) and (2), we get
tan 3θ = (3 tan θ - tan3θ)/(1 - 3tan2θ)
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