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
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 + Sin2 θ = 1
- 1 + Tan2 θ = Sec2 θ
- 1 + Cot2 θ = Cosec2 θ
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 :–
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