Trigonometry Exercise 13.3 – Questions:
I. Show that
- (1 – sin2θ) sec2 θ = 1
- (1 + tan2 θ) cos2 θ = 1
- (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1
- sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ
- 1 + sinθ/1 – sinθ = (sec θ + tan θ)2
- cosA/1-tanA + sinA/1 – cotA = sinA + cosA
- (1 – tan2A)/(1+tan2A) = 1 – 2sin2A
- (sinθ + cosθ)2 = 1 + 2sinθcosθ
- sinA cosA tanA + cosA.sinA.cotA = 1
- (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)
- tan2A – sin2A = tan2A sin2A
- cos2A – sin2A = 2cos2A – 1
Trigonometry Exercise 13.3 – Solutions:
I. Show that
- (1 – sin2θ) sec2 θ = 1
Solution:
(1 – sin2θ) sec2 θ = 1
LHS = (1 – sin2θ) sec2 θ
[since 1 – sin2θ = cos2θ and sec2θ = 1/cos2θ]
= cos2θ .(1/cos2θ)
= 1
= RHS
Thereore, (1 – sin2θ) sec2 θ = 1
- (1 + tan2 θ) cos2 θ = 1
Solution:
(1 + tan2 θ) cos2 θ = 1
LHS = (1 + tan2 θ) cos2 θ [since(1 + tan2 θ) = sec2θ]
= sec2θ.cos2 θ
= (1/cos2θ).cos2 θ
= 1
= RHS
Therefore, (1 + tan2 θ) cos2 θ = 1
- (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1
Solution:
(1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1
LHS = (1 + tan2 θ)(1 – sin θ)( 1 + sin θ)
= (1 + tan2 θ)(1 – sin2 θ) [since (a+b)(a-b) = a2 – b2]
= sec2θ.cos2θ
= (1/cos2θ).cos2θ
= 1
= RHS
Therefore, (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1
- sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ
Solution:
sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ
LHS = sin θ/(1+cosθ) + 1+cosθ/sinθ
= [sinθ.sinθ +(1+cos θ)(1+cos θ)]/[sin θ.(1+cos θ)]
= [sin2 θ+(1+cos θ)2]/[sin θ.(1+cos θ)]
=[sin2 θ + 1+ cos2 θ+2cos θ]/[sin θ(1+cos θ)]
=[2+2cos θ]/[sin θ(1+cos θ)]
=[2(1+cos θ)]/[sin θ(1+cos θ)]
=2/sin θ
= 2 cosec θ
= RHS
Therefore, sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ
- 1 + sinθ/1 – sinθ = (sec θ + tan θ)2
Solution:
1 + sinθ/1 – sinθ = (sec θ + tan θ)2
LHS = 1 + sinθ/1 – sinθ
= 1 + sinθ/1 – sinθ x 1 + sinθ/1 + sinθ
= (1 + sinθ)2/(1- sin2 θ)
= (1 + sinθ)2/(cos2 θ)
= [(1+sinθ)/cosθ]2
= [1/cosθ + sinθ/cosθ]2
= (sec θ + tan θ)2
= RHS
Therefore, 1 + sinθ/1 – sinθ = (sec θ + tan θ)2
- cosA/1-tanA + sinA/1 – cotA = sinA + cosA
Solution:
cosA/1-tanA + sinA/1 – cotA = sinA + cosA
LHS = cosA/1-tanA + sinA/1 – cotA
= cosA/1-(sinA/cosA) + sinA/1 – (cosA/sinA)
= cosA.cosA/cosA-sinA + sinA.sinA/sinA – cosA
= [cos2A/(cosA- sinA)] + [sin2A/(sinA – cosA)]
= [cos2A/(cosA- sinA)] – [sin2A/(cosA – sinA)]
= [cos2A – sin2A]/[cosA – sinA]
= [cosA+sinA][cosA-sinA]/[cosA-sinA]
= cosA + sinA
= RHS
Therefore, cosA/1-tanA + sinA/1 – cotA = sinA + cosA
- (1 – tan2A)/(1+tan2A) = 1 – 2sin2A
Solution:
(1 – tan2A)/(1+tan2A) = 1 – 2sin2A
LHS = (1 – tan2A)/(1+tan2A)
= [1 – (sin2A/cos2A)]/ [1 + (sin2A/cos2A)]
= [(cos2A – sin2A)/cos2A]/[(cos2A + sin2A)/cos2A]
=(cos2A – sin2A)/ (cos2A + sin2A)
= (1 – sin2A – sin2A)/1
= 1 – 2sin2A
= RHS
Therefore, (1 – tan2A)/(1+tan2A) = 1 – 2sin2A
- (sinθ + cosθ)2 = 1 + 2sinθcosθ
Solution:
(sinθ + cosθ)2 = 1 + 2sinθcosθ
LHS = (sinθ + cosθ)2
= sin2 θ + cos2 θ + 2sinθ.cosθ
= 1 + 2sinθcosθ
= RHS
Therefore, (sinθ + cosθ)2 = 1 + 2sinθcosθ
- sinA cosA tanA + cosA.sinA.cotA = 1
Solution:
sinA cosA tanA + cosA.sinA.cotA = 1
LHS = sinA cosA tanA + cosA.sinA.cotA
= sinA cosA sinA/cosA + cosA.sinA. cosA/sinA
= sin2A + cos2A
= 1
= RHS
Therefore, sinA cosA tanA + cosA.sinA.cotA = 1
- (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)
Solution:
(tanA – sinA)/(sin2A) = tanA/ (1 + cosA)
LHS = (tanA – sinA)/(sin2A)
= (sinA/cosA – sinA)/(sin2A)
= [sinA-sinA.cosA]/cosAsin2A
= sinA(1 – cosA)/cosA.(1 – cos2A) [since (a+b)(a-b) = a2 – b2]
= sinA(1 – cosA)/cosA.(1 – cosA)(1 + cosA)
= sinA/cosA (1 + cosA)
= tanA/(1+cosA)
= RHS
Therefore, (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)
- tan2A – sin2A = tan2A sin2A
Solution:
tan2A – sin2A = tan2A sin2A
LHS = tan2A – sin2A
=[sin2A/cos2A] – sin2A
= [(sin2A – sin2A.cos2A)/cos2A] – sin2A
= [sin2A(1 – cos2A)]/cos2A
=sin2A.sin2A/cos2A
= tan2A sin2A
= RHS
Therefore, tan2A – sin2A = tan2A sin2A
- cos2A – sin2A = 2cos2A – 1
Solution:
cos2A – sin2A = 2cos2A – 1
LHS = cos2A – sin2A
= cos2A – (1 – cos2A)
= cos2A – 1 + cos2A)
= 2cos2A – 1
= RHS
Therefore, cos2A – sin2A = 2cos2A – 1
