Trigonometry Exercise 13.3 – Class 10

Trigonometry Exercise 13.3 – Questions:

I. Show that

  1. (1 – sin2θ) sec2 θ = 1
  2. (1 + tan2 θ) cos2 θ = 1
  3. (1 + tan2 θ)(1 – sin θ)( 1 + sin θ) = 1
  4. sin θ/(1+cosθ) + 1+cosθ/sinθ = 2 cosec θ
  5. 1 + sinθ/1 – sinθ = (sec θ + tan θ)2
  6. cosA/1-tanA + sinA/1 – cotA = sinA + cosA
  7. (1 – tan2A)/(1+tan2A) = 1 – 2sin2A
  8. (sinθ + cosθ)2 = 1 + 2sinθcosθ
  9. sinA cosA tanA + cosA.sinA.cotA = 1
  10. (tanA – sinA)/(sin2A) = tanA/ (1 + cosA)
  11. tan2A – sin2A = tan2A sin2A
  12. cos2A – sin2A = 2cos2A – 1

Trigonometry Exercise 13.3 – Solutions:

I. Show that

  1. (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. (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. (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

 

  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. 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

 

  1. 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. (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

 

  1. (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

 

  1. 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

 

  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)

 

  1. 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

 

  1. 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