To learn the important trigonometric identities,
Abbreviations used in the problems :
* L.H.S -----> Left hand side
* R.H.S -----> Right hand side
Problem 1 :
Prove :
cos θ/(sec θ - tan θ) = 1 + sin θ
Solution :
L.H.S :
= cos θ/(sec θ - tan θ)
= cos θ/[1/cos θ - (sin θ/cos θ)]
= cos θ/[(1 - sin θ)/cos θ]
= cos2 θ/(1 - sin θ)
= cos2 θ/(1 - sin θ)
= (1 - sin2 θ)/(1 - sin θ)
= (1 + sin θ) ( 1 - sin θ)/(1 - sin θ)
= 1 + sin θ
= R.H.S
Problem 2 :
Prove :
√(sec2θ + cosec2θ) = tan θ + cot θ
L.H.S :
= √(sec2θ + cosec2θ)
= √(1 + tan2θ + 1 + cot2θ)
= √(2 + tan2θ + cot2θ)
= √(tanθ)2 + (cotθ)2 + 2tanθ x cotθ
= √(tanθ + cotθ)2
= tanθ + cotθ
= R.H.S
Problem 3 :
Prove :
(1 + cos θ - sin2θ)/(sin θ)(1 + cos θ) = cot θ
Solution :
L.H.S :
= (1 + cos θ - sin²θ)/(sin θ)(1+cosθ)
= (1 + cos θ) - (1 - cos²θ)/(sin θ)(1+cosθ)
= [(1 + cos θ) - (1 - cos θ)(1 + cos θ)]/(sin θ)(1+cosθ)
= [(1 + cos θ) (1 - (1 - cos θ))]/(sin θ)(1+cosθ)
taking (1 + cos θ) as common term
= [(1 + cos θ) (1 - 1 + cos θ))]/(sin θ)(1 + cosθ)
= [(1 + cos θ) (cos θ)]/(sin θ)(1 + cosθ)
= cos θ/sin θ
= cot θ
= R.H.S
Problem 4 :
secθ(1 - sinθ)(secθ + tanθ) = 1
Solution :
L.H.S :
Problem 5 :
sinθ/(cosecθ + cotθ) = 1 - cosθ
Solution :
L.H.S :
= sinθ/(cosecθ + cotθ)
= sin θ/[(1/sin θ) + (cos θ/sin θ)]
= sin θ/[(1+cosθ)/sin θ]
= (sin θ x sin θ)/(1+cosθ)
= sin²θ/(1+cosθ)
= 1- cos²θ/(1+cosθ)
= (1- cosθ)(1+cosθ)/(1+cosθ)
= 1 - cos θ
= R.H.S
Problem 6 :
Prove :
[sin(90 - θ)/(1 + sinθ)] + [cosθ/(1 - (cos(90 - θ))] = 2secθ
Solution :
L.H.S :
= [sin(90 - θ)/(1 + sinθ)] + [cosθ/(1 - (cos(90 - θ))]
We can write sin (90-θ) as cos θ and cos (90 - θ) as sin θ.
= 2/cosθ
= 2secθ
= R.H.S
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