Trigonometrical Identities Class 10 RS Aggarwal Exe-22A ICSE Maths Solutions Ch-22. In this article you would learn how to solve problems on trigonometric identities easily. Visit official website CISCE for detail information about ICSE Board Class-10 Mathematics.

Trigonometrical Identities Class 10 RS Aggarwal Exe-22A ICSE Maths Solutions Ch-22

Board	ICSE
Publications	Goyal Brothers Prakashan
Subject	Maths
Class	10th
Chapter-22	Trigonometric Identities
Writer	RS Aggarwal
Book Name	Foundation
Topics	solving trigonometric identities

Trigonometric Identities Problems / Questions with Solutions / Answer

Class 10 RS Aggarwal Exe-22A ICSE Maths Solutions Ch-22

Prove each of the following identities:

Que-1: (cosecA+1)/(cosecA-1) = (1+sinA)/(1-sinA)

Sol: LHS:
cosecA+1/cosecA-1 = (1/sinA)+1/(1/sinA)-1
= (1+sinA)/(sinA) / (1-sinA)/(sinA)
= 1+sinA/1-sinA

LHS=RHS

Que-2: secA-1/secA+1 / 1-cosA/1+cosA

Sol: L.H.S. = sec⁡𝐴−1/sec⁡𝐴+1
= (1/cos⁡𝐴−1/1) /(1/cos⁡𝐴+1/1)
= ((1−cos⁡𝐴)/(cos⁡𝐴))/((1+cos⁡𝐴)/(cos⁡𝐴))
= 1−cos⁡𝐴/cos⁡𝐴 × cos⁡𝐴/1+cos⁡𝐴
= 1−cos⁡𝐴/1+cos⁡𝐴
= R.H.S.

Que-3: sinAtanA/1-cosA = 1+secA

Sol: L.H.S. = sin⁡𝐴⁢tan⁡𝐴/1−cos⁡𝐴
= sin⁡𝐴⁢tan⁡𝐴/1−cos⁡𝐴 × 1+cos⁡𝐴/1+cos⁡𝐴
= sin⁡𝐴⁢tan⁡𝐴⁢(1+cos⁡𝐴)/1−cos²𝐴
= sin⁡𝐴(⁢sin⁡𝐴/cos⁡𝐴)(1+cos⁡𝐴)/sin2⁡𝐴
= 1+cos⁡𝐴/cos⁡𝐴
= 1/cos⁡𝐴 + cos⁡𝐴/cos⁡𝐴
= sec A + 1
= 1 + sec A = R.H.S

Que-4: 1/(tanA+cotA) = cosAsinA

Sol: L.H.S. = 1/tan⁡𝐴+cot⁡𝐴
= 1/(sin⁡𝐴/cos⁡𝐴)+(cos⁡𝐴/sin⁡𝐴)
= 1/(sin²𝐴+cos²𝐴/sin⁡𝐴⁢cos⁡𝐴)
= 1/(1/sin⁡𝐴⁢cos⁡𝐴) …(∵ sin²A + cos²A = 1)
= sin A cos A
= R.H.S.

Que-5: (1+tanA)² + (1-tanA)² = 2sec²A

Sol: L.H.S. = (1 – tan A)² + (1 + tan A)²
= (1 + tan²A – 2 tan A) + (1 + tan²A + 2 tan A)
= 2(1 + tan²A)
= 2 sec²A
= R.H.S.

Que-6: (sin²θ – 1)(tan²θ+ 1) + 1 = 0

Sol: L.H.S. = (sin²θ – 1)(tan²θ + 1) + 1
= (– cos²θ) sec²θ + 1
= −cos²θ ×1/cos²θ +1
= –1 + 1
= 0
= R.H.S.
Hence Proved.

Que-7: cosec(A) (1 + cos A) (cosec A – cot A )=1

Sol: L.H.S. = cosec A(1 + cos A) (cosecA – cot A)
= 1/sin⁡𝐴⁢(1+cos⁡𝐴)⁢(1/sin⁡𝐴−cos⁡𝐴/sin⁡𝐴)
= 1/sin⁡𝐴⁢(1+cos⁡𝐴)⁢(1−cos⁡𝐴/sin⁡𝐴)
= (1+cos⁡𝐴)⁢(1−cos⁡𝐴)/sin²⁡𝐴

Apply the identity (1 + cos⁡A) (1 − cos⁡A) = 1 − cos^⁡2A
=1−cos²𝐴/sin²⁡𝐴
=sin²𝐴/sin²𝐴 =1
cosecA(1 + cosA) (cosecA − cotA) = 1 proved

Que-8: sec(A)  (1 – sin A)(sec(A) + tan A) = 1

Sol: We have to prove sec A(1 − sin A)(sec A + tan A) = 1
We know that sec² A − tan² A − 1
So,
sec A(1 − sin A)(sec A + tan A) = {sec A(1 − sin A)}(sec A + tan A)
= (sec A − sec A sin A)(sec A + tan A)
= (sec⁡𝐴−1/cos⁡𝐴 sin⁡𝐴)⁢(sec⁡𝐴+tan⁡𝐴)     …(∵sec⁡𝜃=1/cos⁡𝜃)
= (sec⁡𝐴−sin⁡𝐴/cos⁡𝐴)⁢(sec⁡𝐴+tan⁡𝐴)    …(∵tan⁡𝜃=sin⁡𝜃/cos⁡𝜃)
= (sec A − tan A)(sec A + tan A)
= sec² A − tan² A
= 1 = R.H.S.    … (∵ sec² θ = 1 tan² θ)

Que-9: (cosec θ-sin θ) (secθ – cos θ)(tan θ+ cot θ) = 1

Sol: Taking LHS
(cosec θ – sinθ )(secθ – cos θ ) ( tanθ +cot θ)
(1/sin⁡𝜃−sin⁡𝜃)⁢(1/cos⁡θ−cos⁡θ)⁢(sin⁡θ/cos⁡θ+cos⁡θ/sin⁡θ)
=(1−sin²θ/sin⁡θ)⁢(1−cos²⁡θ/cos⁡θ)⁢(sin²⁡θ+cos²⁡θ/sin⁡θ.cos⁡θ)
=cos²⁡θ/sin⁡θ ×sin²θ/cos⁡θ ×1/sin⁡θ.cos⁡θ = 1 = RHS

Que-10: (cosec(A) + sin A)(cosec(A) – sin A) = cot² A + cos²A

Sol: L.H.S. = (cosec A + sin A) (cosec A – sin A)
= (cosec² A – sin² A)    …[∵ (a + b) (a – b) = a² – b²]
= 1 + cot² A – sin² A
= cot² A + 1 – sin² A
= cot² A + cos² A  …(∵ 1 – sin² A = cos² A)
= R.H.S.

Que-11: (sec(A) + cos A)(sec(A) – cos A) = sin² A + tan² A

Sol: L.H.S. = (sec A – cos A) (sec A + cos A)
= sec² A – cos² A
= (1 + tan² A) – (1 – sin² A)
= sin² A + tan² A
= R.H.S.

Que-12: tan²A – sin²A = sin²A . tan²A

Sol: LHS:
tan²A − sin²A

We know that:
tan⁡𝐴 =sin⁡𝐴/cos⁡𝐴 ⇒tan²𝐴 =sin²𝐴/cos²⁡𝐴
tan²𝐴 −sin²𝐴 =sin²⁡𝐴/cos²𝐴 −sin²𝐴
=(sin²𝐴−sin²𝐴⁢cos²⁡𝐴)/cos²𝐴

Factor out sin²A:
= sin²𝐴⁢(1−cos²𝐴)/cos²𝐴
1 − cos²A = sin²A
= sin²𝐴⋅sin²𝐴/cos²⁡𝐴 = sin⁴𝐴/cos²⁡𝐴 = sin²A . sin²A/cos²A = tan²A.sin²A
= RHS:
∴LHS=RHS
Hence Proved

Que-13: cot²A – cos²A = cos²A . cot²A

Sol: L.H.S. = cot² A – cos² A
= cos²⁡A/sin²⁡A −cos²A
= (cos²A−sin²A.cos²A)/sin2⁡A
= cos²A⁢(1−sin²⁡A)/sin²⁡A
=  cot² A (cos² A)
= cos² A . cot² A
= R.H.S.

Que-14: sec²A + co sec² A = sec² A . cosec² A

Sol: L.H.S. = sec²A + cosec²A
= 1/cos²𝐴 +1/sin²⁡𝐴
= sin²⁡𝐴+cos²⁡𝐴/cos²𝐴  sin²𝐴
= 1/cos²⁡𝐴 sin²𝐴
= sec²A cosec²A
= R.H.S.   …(∵ sin²A + cos²A = 1)

Que-15: tan²A + cot²A + 2 = sec²A . cosec²A

Sol: In the given question, we need to prove tan² A + cot² A = sec² A cosec² A − 2
Now using tan⁡𝜃 =sin⁡𝜃/cos⁡𝜃 and cot⁡𝜃 =cos⁡𝜃/sin⁡𝜃 in LHS we get
tan²⁡𝐴 +cot²𝐴 =sin²𝐴/cos²⁡𝐴 +cos²𝐴/sin²𝐴
=sin4⁡𝐴+cos4⁡𝐴/cos²𝐴⁢sin²⁡𝐴
=(sin²𝐴)²+(cos²⁡𝐴)²/cos²⁡𝐴⁢sin²⁡𝐴
Further, using the identity 𝑎² +𝑏² =(𝑎+𝑏)² −2⁢𝑎⁢𝑏 we get
(sin²𝐴)²+(cos²𝐴)²/cos²⁡𝐴⁢sin²𝐴 =(sin²𝐴+cos²𝐴)²−2⁢sin²⁡𝐴⁢cos²𝐴/sin²⁡𝐴⁢cos²⁡𝐴
=((1)²−2⁢sin²𝐴⁢cos²𝐴)/sin²𝐴⁢cos²⁡𝐴
=1/sin²⁡𝐴⁢cos²⁡𝐴 −2⁢sin²𝐴⁢cos²𝐴/sin²⁡𝐴⁢cos²⁡𝐴
=cos⁡𝑒⁢𝑐²⁢𝐴⁢sec²⁡𝐴 −2
Since L.H.S = R.H.S
Hence proved.

Que-16: sin A (1 + tan A) + cos A (1 + cot A) = sec(A) + cosec(A)

Sol: L.H.S. = cos A (1 + cot A) + sin A (1 + tan A)
= cos⁡𝐴⁢(1+cos⁡𝐴/sin⁡𝐴) +sin⁡𝐴⁢(1+sin⁡𝐴/cos⁡𝐴)
= cos⁡𝐴⁢(sin⁡𝐴+cos⁡𝐴)/sin⁡𝐴 +sin⁡𝐴⁢(cos⁡𝐴+sin⁡𝐴)/cos⁡𝐴
= (sin⁡𝐴+cos⁡𝐴)⁢[cos⁡𝐴/sin⁡𝐴+sin⁡𝐴/cos⁡𝐴]
= (sin⁡𝐴+cos⁡𝐴)⁢[cos²𝐴+sin²𝐴/sin⁡𝐴⁢cos⁡𝐴]
= (sin⁡𝐴+cos⁡𝐴) ×1/sin⁡𝐴⁢cos⁡𝐴
= sin⁡𝐴+cos⁡𝐴/sin⁡𝐴⁢cos⁡𝐴  …[∵ cos²θ + sin²θ = 1]
= sin⁡𝐴/sin⁡𝐴⁢cos⁡𝐴 +cos⁡𝐴/sin⁡𝐴⁢cos⁡𝐴
= 1/cos⁡𝐴 +1/sin⁡𝐴
= sec A + cosec A = R.H.S.
Since L.H.S = R.H.S
Hence proved.

Que-17: 1/(1 + tan²A) + 1/(1 + cot²A) = 1

Sol: LHS:
1/(1+tan²A) + 1/(1+cot²A)
= 1/(1+tan²A) + 1/(1+(1/tan²A))
= 1/(1+tan²A) + 1/((tan²A+1)/tan²A))
= 1/(1+tan²A) + tan²A/(tan²A+1)
= 1+tan²A/tan²A+1 = 1
=RHS
Since L.H.S = R.H.S
Hence proved.

Que-18: 1/(1 + sin A) + 1/(1 – sin A) = 2sec²A

Sol: L.H.S. = 1/(1−sin⁡𝐴) +1/(1+sin⁡𝐴)
= 1+sin⁡𝐴+1−sin⁡𝐴/(1−sin⁡𝐴)⁢(1+sin⁡𝐴)
= 2/1−sin²𝐴
= 2/cos²⁡𝐴
= 2 sec² A = R.H.S.
Since L.H.S = R.H.S
Hence proved.

Que-19: (1+sinA)/cosA + cosA(1+sinA) = 2secA

Sol: L.H.S. = 1+sin⁡𝐴/cos⁡𝐴 +cos⁡𝐴/1+sin⁡𝐴
= ((1+sin⁡𝐴)²+cos²𝐴)/cos⁡𝐴⁢(1+sin⁡𝐴)
= (1+sin²𝐴+2⁢sin⁡𝐴+cos²⁡𝐴)/cos⁡𝐴⁢(1+sin⁡𝐴)
= (1+2⁢sin⁡𝐴+1)/cos⁡𝐴⁢(1+sin⁡𝐴)
= 2⁢(1+sin⁡𝐴)/cos⁡𝐴⁢(1+sin⁡𝐴)
= 2 sec A = R.H.S

Que-20: cosecA/(cosecA-1) + cosecA/(cosecA+1) = 2sec²A

Sol: L.H.S.  =cos⁡𝑒⁢𝑐⁢𝐴/cos⁡𝑒⁢𝑐⁢𝐴−1 +cos⁡𝑒⁢𝑐⁢𝐴/cos⁡𝑒⁢𝑐⁢𝐴+1
= (cos⁡𝑒⁢𝑐⁢𝐴⁢(cos⁡𝑒⁢𝑐⁢𝐴+1)+cos⁡𝑒⁢𝑐⁢𝐴⁢(cos⁡𝑒⁢𝑐⁢𝐴−1))/(cos⁡𝑒⁢𝑐⁢𝐴−1)⁢(cos⁡𝑒⁢𝑐⁢𝐴+1)
= (cos⁡𝑒⁢𝑐²𝐴+cos⁡𝑒⁢𝑐 𝐴+cos⁡𝑒⁢𝑐²⁢𝐴−cos⁡𝑒⁢𝑐 𝐴)/(cos⁡𝑒⁢𝑐⁢𝐴)²−(1)²
= 2⁢cos⁡𝑒⁢𝑐²⁢𝐴/cos⁡𝑒⁢𝑐²𝐴−1
= 2⁢cos⁡𝑒⁢𝑐²𝐴/cot²𝐴    …(∵ cosec² A – 1 = cot² A)
= 2(⁢(1/sin²⁡𝐴)/(cos²⁡𝐴/sin²𝐴))
= 2/cos²𝐴
= 2 sec² A
= R.H.S.

Que-21: (1+cotA-cosecA) (1+tanA+secA) = 2

 Sol: L.H.S:(1+cos⁡𝐴/sin⁡𝐴−1sin⁡𝐴)⁢(1+sin⁡𝐴/cos⁡𝐴+1/cos⁡𝐴)
=(sin⁡𝐴+cos⁡𝐴−1/sin⁡𝐴)⁢(cos⁡𝐴+sin⁡𝐴+1/cos⁡𝐴)
=(sin⁡𝐴+cos⁡𝐴)²−(1)²/sin⁡𝐴.cos⁡𝐴
=sin²𝐴+cos²⁡𝐴+2⁢sin⁡𝐴.cos⁡𝐴−1/sin⁡𝐴.cos⁡𝐴
=1+2⁢sin⁡𝐴.cos⁡𝐴−1/sin⁡𝐴.cos⁡𝐴
=2
Hence, L.H.S =R.H.S.

Que-22: (sinA+cosecA)² + (cosA+secA)² = 7 + tan²A + cot²A

Sol: L.H.S. = (sin A + cosec A)² + (cos A + sec A)²
= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
= sin2⁡𝐴 +cos⁡𝑒⁢𝑐2⁢𝐴 +2⁢sin⁡𝐴 ×1/sin⁡𝐴 +cos2⁡𝐴 +sec2⁡𝐴 +2⁢cos⁡𝐴 ×1cos⁡𝐴
= sin² A + cos² A + cosec² A + sec² A + 2 + 2  …(∵ sin² A + cos² A = 1)
= 1 + cosec² A + sec² A + 4
= (1 + cot² A) + (1 + tan² A) + 5  …[∵ cosec² A = 1 + cot² A and sec² A = 1 + tan² A]
= 1 + cot² A + 1 + tan² A + 5
= 7 + tan² A + cot² A
= R.H.S.

Que-23: cos³A+sin³A/cosA+sinA + cos³A-sin³A/cosA-sinA = 2

Sol: LHS:
cos³A+sin³A/cosA+sinA + cos³A-sin³A/cosA-sinA
= ((cosA-sinA)(cos³A+sin³A) + (cosA+sinA)(cos³A-sin³A))/(cosA+sinA)(cosA-sinA)
= (2cos⁴A – 2sin⁴A)/(cos²A-sin²A)
= 2((cos²A)² – (sin²A)²)/(cos²A-sin²A)
= 2(cos²A-sin²A)(cos²A+sin²A)/(cos²A-sin²A)
= 2(cos²A+sin²A)
= 2 × 1
= 2
= RHS

Que-24: tanA/(1-cotA) + cotA/(1-tanA) = secA cosecA + 1

Sol: LHS:
tanA/(1-cotA) + cotA/(1-tanA)
= (sinA/cosA)(1-cosA/sinA) + (cosA/sinA)(1-sinA/cosA)
= (sinA/cosA)((sinA-cosA)/sinA) + (cosA/sinA)((cosA-sinA)/cosA)
= sin²A/cosA(sinA-cosA) + cos²A/sinA(cosA-sinA)
= sin²A/cosA(sinA-cosA) – cos²A/sinA(sinA-cosA)
= (sin³A – cos³A)/sinA.cosA(sinA-cosA)
= (sinA-cosA)(sin²A+cos²A+sinA.cosA)/sinA.cosA(sinA-cosA)
= (sin²A+cos²A+sinA.cosA)/sinA.cosA
= (1+sinA.cosA)/sinA.cosA
= 1/sinA.cosA + sinAcosA/sinA.cosA
= secA.cosecA + 1
= RHS

Que-25: sinA/(1+cotA) – cosA/(1+tanA) = sinA – cosA

Sol: LHS:
sinA/(1+cotA) – cosA/(1+tanA)
= sinA/(1+(cosA/sinA)) – cosA/(1+(sinA/cosA))
= sinA/((sinA+cosA)/(sinA)) – cosA/((cosA+sinA)/(cosA))
= sin²A/(sinA+cosA) – cos²A/(cosA+sinA)
= (sin²A-cos²A)/(sinA+cosA)
= (sinA-cosA)(sinA+cosA)/(sinA+cosA)
= sinA-cosA
= RHS

Que-26: tanθ+sinθ/tanθ-sinθ = secθ+1/secθ-1

Sol: LHS:
tanθ+sinθ/tanθ-sinθ
= (sinθ/cosθ)+sinθ/(sinθ/cosθ)-sinθ
= sinθ(1/cosθ+1)/sinθ(1/cosθ-1)
= secθ+1/secθ-1
= RHS

Que-27: cotθ+cosecθ-1/cotθ-cosecθ+1 = 1+cosθ/sinθ

Sol: LHS:
cotθ+cosecθ-1/cotθ-cosecθ+1 = 1+cosθ/sinθ
= (cotθ+cosecθ)-(cosec²θ-cot²θ)/(cotθ-cosecθ+1) [cot²θ – cosec²θ = 1]
= (cotθ+cosecθ)-(cosecθ-cotθ)(cosecθ+cotθ)/(cotθ-cosecθ+1)
= (cotθ+cosecθ)-(1-cosecθ+cotθ)/(cotθ-cosecθ+1)
= (cotθ+cosecθ)
= (cosθ/sinθ)+(1/sinθ)
= (1+cosθ)/sinθ
= RHS

Que-28: (cotA-1)/(2-sec²A) = cotA/(1+tanA)

Sol: LHS:
((1-tanA)/tanA) / (2-(1+tan²A))
= ((1-tanA)/tanA) / (1-tan²A)
= ((1-tanA)/tanA) / (1+tanA)(1-tanA)
= (1/tanA)/(1+tanA)
= cotA/(1+tanA)
= RHS

Que-29: 1/(secA+tanA) – 1/cosA = 1/cosA – 1/(secA-tanA)

Sol: LHS:
1/(secA+tanA) – 1/cosA
= (sec²A-tan²A)/(secA+tanA) – 1/cosA
= (secA+tanA)(secA-tanA)/(secA+tanA) – 1/cosA
= (secA-tanA) – secA
= secA – tanA – secA
= secA – (tanA+secA)
= secA – (tanA+secA)/(sec²A-tan²A)
= secA – (tanA+secA)/(secA+tanA)(secA-tanA)
= secA – 1/(secA-tanA)
= 1/cosA – 1/(secA-tanA)
= RHS

Que-30: sinA/(cotA+cosecA) = 2 + sinA/(cotA-cosecA)

Sol: LHS:
sinA/(cotA+cosecA) = 2 + sinA/(cotA-cosecA)
sinA/(cotA+cosecA) – sinA/(cotA-cosecA) = 2
LHS:
sinA(1/(cotA+cosecA) + 1/(cosecA-cotA))
= sinA(cosecA-cotA + (cotA+cosecA))/(cosec²A-cot²A)
= sinA(cosecA-cotA + cotA+cosecA)/(cosec²A-cot²A)
= sinA(2cosecA)/(cosec²A-cot²A)
= sinA(2cosecA)/1
= 2 × sinA × cosecA
= 2 × sinA × 1/sinA
= 2
= RHS

Que-31: cotA-tanA = 2cos²A-1/sinA.cosA

Sol: LHS:
cotA – tanA
= cosA/sinA –  sinA/cosA
= (cos²A-sin²A)/sinA.cosA
= (cos²A-(1-cos²A))/sinA.cosA
= (2cos²A-1)/sinA.cosA
= RHS

Que-32: √(1+cosA/1-cosA) = (cosecA-cotA)

Sol: LHS:
√(1+cosA/1-cosA)
= √(1+cosA/1-cosA)×(1-cosA/1-cosA)
= √(1-cosA)²/1-cos²A
= √(1-cosA)²/sin²A
= 1-cosA/sinA
= 1/sinA – cosA/sinA
= cosecA – cotA
= RHS

Que-33: √(1-sinA/1+sinA) = (secA-tanA)

Sol: L.H.S.
= √(1+sin⁡𝐴/1−sin⁡𝐴)
= √(1+sin⁡𝐴/1−sin⁡𝐴)×(1+sin⁡𝐴/1+sin⁡𝐴)
= √(1+sin⁡𝐴)²/(1−sin²⁡𝐴)
= √(1+sin⁡𝐴)²/cos²⁡𝐴
= (1+sin⁡𝐴)/cos⁡𝐴
= 1/cos⁡𝐴 +sin⁡𝐴/cos⁡𝐴
= sec A + tan A
= R.H.S.

Que-34: cot²A/(cosecA+1)² = 1-sinA/1+sinA

Sol: RHS:
= 1-sinA/1+sinA
= 1-(1/cosecA)/1+(1/cosecA)
= cosecA-1/cosecA+1
= (cosecA-1/cosecA+1)×(cosecA+1/cosecA+1)
= (cosec²A-1)/(cosecA+1)²
= cot²A/(cosecA+1)²
= LHS

Que-35: cosA/(1-tanA) + sin²A/(sinA-cosA) = (cosA+sinA)

Sol: LHS:
= cosA/(1-sinA/cosA) + sin²A/(sinA-cosA)
= cosA/(cosA-sinA/cosA) + sin²A/(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

Que-36: (1-tanθ/1-cotθ)² = tan²θ

Sol: LHS:
(1-tanθ/1-cotθ)²
= (1-tanθ/1-(1/tanθ))²
= (1-tanθ/(tanθ-1)/tanθ)²
= (-tanθ)²
= tan²θ
= RHS

Que-37: cosA/(1+sinA) + tanA = secA

Sol: LHS:
cosA/(1+sinA) + tanA
= cosA(1-sinA)/(1+sinA)(1-sinA) + sinA/cosA
= cosA-sinA.cosA/(1-sin²A) + sinA/cosA
= cosA-sinA.cosA/cos²A + sinA/cosA
= 1/cosA – sinA/cosA + sinA/cosA
= 1/cosA
= RHS

Que-38: (sinθ + cosθ)(tanθ + cotθ) = secθ + cosecθ

Sol: LHS:
(sinθ + cosθ)(tanθ + cotθ)
= (sinθ+cosθ)(sinθ/cosθ + cosθ/sinθ)
= (sinθ+cosθ)(sin²θ+cos²θ/sinθ.cosθ)
= (sinθ+cosθ) × 1/sinθ.cosθ
= (sinθ+cosθ)/sinθ.cosθ
= sinθ/sinθ.cosθ + cosθ/sinθ.cosθ
= 1/cosθ + 1/sinθ
= secθ + cosecθ
= RHS

Que-39: (1+tan²A) + (1+cot²A) = 1/(sin²A-sin⁴A)

Sol: LHS:
(1+tan²A) + (1+cot²A)
= (1+sin²A/cos²A) + (1+cos²A/sin²A)
= (cos²A+sin²A/cos²A) + (sin²A+cos²A/sin²A)
= 1/cos²A + 1/sin²A
= 1/(1-sin²A) + 1/sin²A
= sin²A+1-sin²A/sin²A(1-sin²A) = 1/sin²A-sin⁴A
= RHS

Que-40: √(sec²A+cosec²A) = (tanA+cotA)

Sol: L.H.S. = √(sec²⁡𝐴+cos⁡𝑒⁢𝑐²𝐴)
= √(1/cos²𝐴+1/sin²⁡𝐴)
= √(sin²⁡𝐴+cos²⁡𝐴)/(sin²⁡𝐴⁢cos²⁡𝐴)
= √1/(sin²⁡𝐴⁢cos²𝐴)
= √1/(sin⁡𝐴⁢cos⁡𝐴)

R.H.S.
= tan A + cot A
= sin⁡𝐴/cos⁡𝐴 +cos⁡𝐴/sin⁡𝐴
= (sin²𝐴+cos²⁡𝐴)/(sin⁡𝐴⁢cos⁡𝐴)
= 1/(sin⁡𝐴⁢cos⁡𝐴)

L.H.S. = R.H.S.

Que-41: (tanA+cotA)(cosecA-sinA)(secA-cosA) = 1

Sol: LHS:
(tanA+cotA)(cosecA-sinA)(secA-cosA)
= (sinA/cosA + cosA/sinA)(1/sinA – sinA)(1/cosA – cosA)
= (sin²A+cos²A/sinA.cosA)(1-sin²A/sinA)(1-cos²A/cosA)
= (1/sinA.cosA)(cos²A/sinA)(sin²A/cosA)
= sin²A.cos²A/sin²A.cos²A
= 1

Que-42: (1+sinθ)²(1-sinθ)²/2cos²θ = sec²θ+tan²θ

Sol: LHS:
(1+sinθ)²(1-sinθ)²/2cos²θ
= (1⁢+sin²⁡𝜃+2⁢sin⁡𝜃+1+sin²𝜃− 2⁢sin⁡𝜃)/2⁢cos⁡𝜃
⇒2⁢(1+sin²⁡𝜃)/2⁢cos²𝜃
= (1+sin²𝜃)/cos²𝜃 = 1/cos²𝜃 + sin²𝜃/cos²𝜃
= sec²𝜃 + tan²𝜃
= RHS

— : End of Trigonometrical Identities Class 10 RS Aggarwal Exe-22A ICSE Maths Ch-22 Practice Questions :–

Return to:-  ICSE Class 10 Maths RS Aggarwal Solutions
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