Trigonometrical Identities and Tables Class 10 OP Malhotra Exe-16A ICSE Maths Solutions

Trigonometrical Identities and Tables Class 10 OP Malhotra Exe-16A ICSE Maths Solutions Ch-16 questions as latest prescribe guideline for upcoming exam. In this article you would learn Proving Trigonometrical Identities easily. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.

Trigonometrical Identities and Tables Class 10 OP Malhotra Exe-16A ICSE Maths Solutions

Trigonometrical Identities and Tables Class 10 OP Malhotra Exe-16A ICSE Maths Solutions

Board ICSE
Publications  S Chand
Subject Maths
Class 10th
Chapter-16 Trigonometrical Identities and Tables
Writer OP Malhotra
Exe-16A Proving Trigonometrical Identities
Edition 2024-2025

How to prove Trigonometrical Identities Easily

The general method of proving trigonometric identities is to work on each side of the equation separately, and simplify or manipulate each side until you reach the same expression on both sides. We’re done once we’ve reached the same expression on both sides of the equation. To prove any identity use following identity

  • Cos2 θ + Sinθ = 1
  • 1 + Tanθ = Secθ
  • 1 + Cotθ = Cosecθ

The best way to become better and faster at verifying trigonometric identities is to do practice problems. The more and more problems that you do, the better you will become at recognizing which strategies to employ in solving those problem

Exercise- 16A (Proving Identities)

Trigonometrical Identities and Tables Class 10 OP Malhotra Exe-16A ICSE Maths Solutions Ch-15

Prove that:

Que-1: {1−cos²θ}/sin²θ = 1.

Sol: L.H.S. = {1−cos²θ}/sin²θ
= sin²θ/sin²θ
= 1 = R.H.S. {∵ 1 – cos² θ = sin² θ}

Que-2: {1−sin²θ}/cos²θ = 1.

Sol: L.H.S. = {1−sin²θ}/cos²θ
= cos²θ/cos²θ
= 1 = R.H.S. {∵ 1 – sin² θ = cos² θ}

Que-3: sin A . cot A = cos A.

Sol: L.H.S. = sin A . cot A
= sin A = cosA/sinA
= cos A = R.H.S.

Que-4: {1/cos²θ} − tan²θ = 1

Sol: L.H.S. = {1/cos²θ} − tan²θ
= sec² θ – tan² θ
= 1 = R.H.S.

Que-5: tan2 A cos2 A = 1 – cos2 A.

Sol: L.H.S. = tan² A cos² A = {sin²A/cos²A}cos²A
= sin2 A = 1 – cos2 A = R.H.S. {∵sin2 A = 1 – cos2 A}

Que-6: tan θ = sinθ/√{1−sin²θ}

Sol: R.H.S. = sinθ/√{1−sin²θ} = sinθ/√cos²θ {∵ 1 – sin² θ = cos² θ}
= sinθ/cosθ = tan θ = L.H.S.

Que-7: (1+cosθ)/sin²θ = 1/(1−cosθ)

Sol: L.H.S. = (1+cosθ)/sin²θ = (1+cosθ)/(1−cos²θ) {∵ sin² θ = 1 – cos² θ}
= (1+cosθ)/{(1+cosθ)(1−cosθ)} {∵ a² – b² = (a + b)(a – b)}
= 1/(1−cosθ) = R.H.S.

Que-8: cot2 θ (1 – cos2 θ) = cos2 θ.

Sol: L.H.S. = cot² θ (1 – cos² θ)
= (cos²θ/sin²θ) × sin²θ
= cos²θ = R.H.S.

Que-9: tan2 θ (1 – sin2 θ) = sin2 θ

Sol: L.H.S. = tan² θ (1 – sin² θ)
= (sin²θ/cos²θ) × cos²θ
= sin²θ = R.H.S.

Que-10: (1 – sin2 θ) sec2 θ = 1.

Sol: L.H.S. = (1 – sin² θ) sec² θ
= cos² θ × (1/cos²θ)
= 1 = R.H.S.

Que-11: (1 – cos2 θ) cosec2 θ = 1.

Sol: L.H.S. = (1 – sin² θ) sec² θ
= sin² θ × (1/sin²θ)
= 1 = R.H.S.

Que-12: sin² θ + {1/(1+tan²θ)} = 1.

Sol: L.H.S. = sin² θ + {1/(1+tan²θ)}
= sin² θ + (1/sec²θ) {∵ 1 + tan² θ = sec² θ}
= sin² θ + cos² θ
= 1 = R.H.S.  { ∵ sin2 θ + cos2 θ = 1}

Que-13 cos²θ + {1/(1+cot²θ)} = 1

Sol:  L.H.S. = cos² θ + {1/(1+cot²θ)}
= cos² θ + (1/cosec²θ) {∵ 1 + cot² θ = cosec² θ}
= cos² θ + sin² θ
= 1 = R.H.S.  { ∵ sin2 θ + cos2 θ = 1}

Que-14: {sin²θ+cos²θ}/{sec²θ−tan²θ} = 1

Sol: L.H.S. = {sin²θ+cos²θ}/{sec²θ−tan²θ} = 1/1
{ ∵ sin² θ + cos² θ = 1 and sec² θ – tan² θ = 1}
= 1 = R.H.S.

Que-15: {(cos²A/sin²A)+1} tan² A = 1/cos²A

Sol: L.H.S. = {(cos²A/sin²A)+1} tan² A
= (cot2 A + 1) tan2 A
= cosec2 A × tan2 A
= {1/sin²A} × {sin²A/cos²A}
= 1/cos²A = R.H.S.

Que-16: sin4 θ + sin2 θ cos2 θ = sin2 θ.

Sol: L.H.S. = sin4 θ + sin2 θ cos2 θ
= sin2 θ(sin2 θ + cos2 θ) {∵ sin2 θ + cos2 θ = 1}
= sin2 θ × 1 = sin2 θ = R.H.S.

Que-17: sin4 θ + 2 sin2 θ cos2 θ + cos4 θ = 1.

Sol: L.H.S. = sin4 θ + 2 sin2 θ cos2 θ + cos4 θ
= (sin2 θ)2 + 2 sin2 θ cos2 θ + (cos2 θ)2
= (sin2 θ + cos2 θ)2
= (1)2 = 1 = R.H.S.

Que-18: sin4 A cosec2 A + cos4 A sec2 A = 1.

Sol: L.H.S. = sin4 A cosec2 A + cos4 A sec2 A
= sin4 A × (1/sin2A) + cos4 A × (1/cos2A)
= sin2 θ + cos2 θ = 1 = R.H.S. {∵ sin2 θ + cos2 θ = 1}

Que-19: sin2 A cot2 A + cos2 A tan2 A = 1.

Sol: L.H.S. = sin² A cot² A + cos² A tan² A
= sin² A × (cos²A/sin²A) + cos² A × (sin²A/cos²A)
= sin²A + cos²A
= 1 = R.H.S.

Que-20: tan θ + cot θ = sec θ cosec θ

Sol: L.H.S. = tan θ + cot θ = (sinθ/cosθ) + (cosθ/sinθ)
= (sin²θ+cos²θ)/(cosθsinθ)
= 1/(cosθsinθ)
= secθ cosecθ = R.H.S  {1/cosθ = secθ   or  1/sinθ = cosecθ}

Que-21: (tan A + cot A) sin A cos A = 1

Sol: L.H.S. = (tan A + cot A) (sin A cos A)
= (sinA/cosA + cosA/sinA) sin A cosA
= {(sin²A+cos²A)/cosAsinA} × sin A cos A
= {1/cosA sinA} × sin A cos A { ∵ sin2θ + cos2 θ = 1}
= 1 = R.H.S.

Que-22: {1+cosθ−sin²θ}/{sinθ(1+cosθ)} = cotθ

Sol: L.H.S. = {1+cosθ−sin²θ}/{sinθ(1+cosθ)}
= {1−sin²θ+cosθ}/{sinθ(1+cosθ)} {∵ 1 – sin2 θ = cos2 θ}
= {cosθ+cos²θ}/{sinθ(1+cosθ)}
= {cosθ(1+cosθ)}/{sinθ(1+cosθ)}
= cosθ/sinθ = cot θ = R.H.S.

Que-23: {1/(1−cosθ)} + {1/(1+cosθ)} = 2 cosec² θ

Sol: L.H.S. = {1/(1−cosθ)} + {1/(1+cosθ)
= {(1+cosθ)+(1−cosθ)/{(1−cosθ)(1+cosθ)}
= 2/(1−cos²θ)    {(a + b) (a – b) = a² – b²}
= 2/sin²θ = 2 cosec² θ

Que-24: {1−tan²θ}/{cot²θ−1} = tan²θ

Sol: L.H.S. = {1−tan²θ}/{cot²θ−1}
= {1−tan²θ}/{(1/tan²θ)-1}
= {1−tan²θ}/[{{1−tan²θ}}/{tan²θ}]
= tan²θ = R.H.S.

Que-25: 1/(secθ+tanθ) = (1−sinθ)/cosθ

Sol: L.H.S. = 1/(secθ+tanθ)
= (secθ-tanθ)/{(secθ+tanθ)(secθ-tanθ)}
= (secθ-tanθ)/(sec²θ-tan²θ)
= (secθ-tanθ)/1   {sec²θ-tan²θ = 1}
= (1/cosθ) – (sinθ/cosθ)
= (1-cosθ)/sinθ = R.H.S.

Que-26: (cosec A – sin A) (sec A – cos A) (tan A + cot A ) = 1.

Sol: L.H.S = (cosec A – sin A) (sec A – cos A) (tan A + cot A )
= {(1/sinA)-sinA} {(1/cosA)-cosA} {(sinA/cosA)+(cosA/sinA)}
= {(1-sin²A)/sinA} {(1-cos²A)/cosA} {(sin²A+cos²A)/(sinA cosA)}
= {cos²A/sinA} {sin²A/cosA} {1/(sinA cosA)}
= {sin²A cos²A}/{cos²A sin²A}
= 1 = R.H.S.

Que-27: (1+sinθ)/(1−sinθ) = (sec θ + tan θ)²

Sol: L.H.S. = (1+sinθ)/(1−sinθ)
= {(1+sinθ)(1+sinθ)}/{(1−sinθ)(1+sinθ)}
= {(1+sinθ)²}/{1−sin²θ}
= {(1+sinθ)²/cos²θ (∵ 1 – sin² θ = cos² θ)
= {(1+sinθ)/cosθ}²
= {(1/cosθ)+(sinθ/cosθ)}²
= (sec θ + tan θ)² = R.H.S.

Que-28: (1+cosθ)/(1−cosθ) = (cosec θ + cot θ)²

Sol: L.H.S. = (1+cosθ)/(1−cosθ)
= {(1+cosθ)(1+cosθ)}/{(1−cosθ)(1+cosθ)}
= {(1+cosθ)²}/{1−cos²θ}
= {(1+cosθ)²/sin²θ (∵ 1 – cos² θ = sin² θ)
= {(1+cosθ)/sinθ}²
= {(1/sinθ)+(cosθ/sinθ)}²
= (cosec θ + cot θ)² = R.H.S.

Que-29: {cotθ+cosecθ-1}/{cotθ-cosecθ+1} = cosecθ+cotθ = {(1+cosθ)/sinθ}

Sol: L.H.S = {cotθ+cosecθ-1}/{cotθ-cosecθ+1}
= {cotθ+cosecθ-(cosec²θ-cot²θ)}/{cotθ-cosecθ+1}
= {cotθ+cosecθ+(cot²θ-cosec²θ)}/{cotθ-cosecθ+1}
= {(cotθ+cosecθ)+(cotθ+cosecθ)(cotθ-cosecθ)}/{cotθ-cosecθ+1}
= {(cotθ+cosecθ)(1+cotθ-cosecθ)}/{cotθ-cosecθ+1}
= {(cotθ+cosecθ)(cotθ-cosecθ+1)}/{cotθ-cosecθ+1}
= cotθ + cosecθ = cosecθ+cotθ
= (1/sinθ) + (cosθ/sinθ)
= (1+cosθ)/sinθ = R.H.S.

Que-30: If tan θ + cot θ = 2, Prove that tan2 θ + cot2 θ = 2.

Sol: tan θ + cot θ = 2
Squaring both sides,
(tan θ + cot θ)2 = 4
tan2 θ + cot2 θ + 2 tan θ cot θ = 4
⇒ tan2 θ + cot2 θ + 2 × 1 = 4 { ∵ tan θ cot θ = 1}
⇒ tan2 θ + cot2 θ + 2 = 4
⇒ tan2 θ + cot2 θ = 4 – 2 = 2
∴ tan2 θ + cot2 θ = 2

Que-31: If sin θ + cos θ = a, sin θ – cos θ = b, Prove that a2 + b2 = 2

Sol: a2 + b2 = (sin θ + cos θ)2 + (sin θ – cos θ)2
= sin2 θ + cos2 θ + 2 sin θ cos θ + sin2 θ + cos2 θ – 2sin θ cos θ
= 2 sin2 θ + 2 cos2 θ = 2 (sin2 θ + cos2 θ)
= 2 × 1 = 2 {∵ sin2 θ + cos2 θ = 1} Hence Proved.

–: End of Trigonometrical Identities and Tables Class 10 OP Malhotra Exe-16A ICSE Maths Solutions :–

Return to :  OP Malhotra S Chand Solutions for ICSE Class-10 Maths

Thanks

Please Share with Your Friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.

error: Content is protected !!