Trigonometrical Identities Class 10 Concise Exe-21A ICSE Maths Selina Solutions Ch-21. In this article you would learn how to solve problems / questions on Trigonometric Ratios and Identities. Visit official website CISCE for detail information about ICSE Board Class-10 Mathematics.

Trigonometrical Identities Class 10 Concise Exe-21A ICSE Maths Selina Solutions Ch-21

Board	ICSE
Publications	Selina
Subject	Maths
Class	10th
Chapter-21	Trigonometrical Identities
Writer	R.K. Bansal
Exe-21A	Trigonometric Ratios and Identities.
Edition	2025-2026

Trigonometric Ratios and Identities

Class 10 Concise Exe-21A ICSE Maths Selina Solutions Ch-21

Prove the following identities :

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

Sol: RHS = (cosecA+1)/(cosecA-1)
= {(1/sinA)+1} / {(1/sinA)-1}
= (1+sinA)/(1-sinA)
= L.H.S

Que-3: 1/(tanA + cotA) = cosA sinA

Sol: L.H.S. = 1/(tan⁡𝐴+cot⁡𝐴)
= 1/{(sin⁡𝐴/cos⁡𝐴)+(cos⁡𝐴/sin⁡𝐴)}
= 1/{(sin²⁡𝐴+cos²⁡𝐴)/(sin⁡𝐴⁢cos⁡𝐴)}
= 1/{1/(sin⁡𝐴⁢cos⁡𝐴)}   …(∵ sin²A + cos²A = 1)
= sin A cos A
= R.H.S.

Que-4: tan A – cot A = (1-2cos²A)/(sinA cosA)

Sol:  tan⁡𝐴 −cot⁡𝐴 = (sin⁡𝐴/cos⁡𝐴) − (cos⁡𝐴/sin⁡𝐴)
= (sin²⁡𝐴−cos²⁡𝐴)/(sin⁡𝐴⁢cos⁡𝐴)
= {1−cos²⁡𝐴−cos²⁡𝐴}/(sin⁡𝐴⁢cos⁡𝐴)  (𝑄⁢ sin²𝐴 = 1 − cos²⁡𝐴)
= (1 − 2⁢cos²⁡𝐴)/(sin⁡𝐴⁢cos⁡𝐴)

Que-5: (sin⁴ A – cos⁴ A) = sin² A – 1

Sol:  L.H.S. = sin⁴A – cos⁴A
= (sin²A)² – (cos²A)²
= (sin²A + cos²A)(sin²A – cos²A)
= sin²A – cos²A
= sin²A – (1 – sin²A)
= 2sin²A – 1
= R.H.S.

Que-6: (1-tanA)² + (1+tanA)² = 2 sec² A

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-7: cosec⁴ A – cosec² A = cot⁴ A + cot² A

Sol: L.H.S. = cosec⁴ A – cosec² A
= cosec² A (cosec² A – 1)
R.H.S. = cot⁴ A + cot² A
= cot² A (cot² A + 1)
= (cosec² A – 1) cosec² A
Thus, L.H.S. = R.H.S.

Que-8: secA (1-sinA) (secA+tanA) = 1

Sol: LHS = secA(1 – sinA) (sec A + tan A)
= (1/cos⁡𝐴) ⁢(1−sin ⁡𝐴) ⁢{(1/cos⁡𝐴) + (sin⁡𝐴/cos⁡𝐴)
= {(1−sin⁡𝐴)/cos⁡𝐴} {⁢(1+sin⁡𝐴)/cos⁡𝐴} = {(1−sin²𝐴)/cos²⁡𝐴}
= (cos²⁡𝐴) / (cos²𝐴)
= 1 = RHS

Que-9: cosec A (1 + cos A) (cosec A – cot A) = 1

Sol:  LHS = cosec A (1 + cos A) (cosec A – cot A)

Que-10: sec² A + cosec² A = sec² A . cosec² A

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

Que-11: {(1+tan² A) cotA} / cosec² A = tan A

Sol: L.H.S. = {(1+tan²⁡𝐴)⁢cot⁡𝐴} / cos⁡𝑒⁢𝑐²⁢ 𝐴
= {(sec²𝐴)⁢cot⁡𝐴}/cos⁡𝑒⁢𝑐²⁢𝐴  …(∵ sec² A = 1 + tan² A)
= {(1/cos²𝐴)×(cos⁡𝐴/sin⁡𝐴)} / {1/sin²⁡𝐴}
= {1/(cos⁡𝐴⁢sin⁡𝐴)} / (1/sin²⁡𝐴)
= sin⁡𝐴/cos⁡𝐴
= tan A = R.H.S.

Que-12: tan² A – sin² A = tan² A . sin² A 

Sol: L.H.S. = tan² A – sin² A
= (sin²𝐴/cos²⁡𝐴) − (sin²⁡𝐴/1)
= (sin²⁡𝐴−sin²⁡𝐴⁢cos²𝐴)/cos²⁡𝐴
= {sin²⁡𝐴⁢(1−cos²⁡𝐴)}/cos²⁡𝐴
= (sin²⁡𝐴/cos²𝐴) ⁢(1−cos²𝐴)
= tan² A sin² A   …(∵ 1 – cos² A = sin² A)
= R.H.S.

Que-13: cot² A – cos² A = cot² A . cos² A

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

Que-14: (cosec A + sin A) (cosec A – sin A) = cot² A + cos² A

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-15: (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-16: (cos A + sin A)² + (cos A – sin A)² = 2

Sol: L.H.S. = (cos A + sin A)² + (cos A – sin A)²
= cos² A + sin² A + 2 cos A . sin A + cos² A + sin² A – 2 cos A . sin A
= 2 sin² A + 2 cos² A
= 2(sin² A + cos² A)   …(∵ sin² A + cos² A = 1)
= 2 × 1
= 2
= R.H.S

Que-17: (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1

Sol: LHS = (cosec A – sin A) (sec A – cos A) (tan A + cot A)

= 1

Que-18: 1/(sec A + tan A) = sec A – tan A

Sol: L.H.S. = 1/(sec⁡𝐴+tan⁡𝐴)
= 1/{(1/cos⁡𝐴)+(sin⁡𝐴/cos⁡𝐴)}
= 1 / {(1+sin⁡𝐴)/cos⁡𝐴}
= {cos⁡𝐴/(1+sin⁡𝐴)} × {(1−sin⁡𝐴)/(1+sin⁡𝐴)}
= {cos⁡𝐴⁢(1−sin⁡𝐴)} / {(1)²−sin²⁡𝐴}
= {cos⁡𝐴⁢(1−sin⁡𝐴)} / cos²⁡𝐴
= (1/cos⁡𝐴) − (sin⁡𝐴/cos⁡𝐴)
= sec A – tan A
L.H.S. = R.H.S.
Hence proved.

Que-19: cosec A + cot A = 1/(cosec A – cot A)

Sol: L.H.S. = cos⁡𝑒⁢𝑐⁢𝐴 + cot⁡𝐴
= {(cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴)/1} × {(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)/(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)}
= {cos⁡𝑒⁢𝑐²⁢𝐴−cot²𝐴} / (cos⁡𝑒⁢𝑐⁢𝐴−cot⁡A)
= {1+cot²⁡𝐴−cot²⁡𝐴} / (cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)
= 1/(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)
= R.H.S.

Que-20: (secA-tanA)/(secA+tanA) = 1 – 2 sec A tan A + 2tan²A

Sol: L.H.S. = (sec⁡𝐴−tan⁡𝐴)/(sec⁡𝐴+tan⁡𝐴)
= {(sec⁡𝐴−tan⁡𝐴)/(sec⁡𝐴+tan⁡𝐴)} × {(sec⁡𝐴−tan⁡𝐴)/(sec⁡𝐴−tan⁡𝐴)}
= {(sec⁡𝐴−tan⁡𝐴)²⁡}/(sec²⁡⁡𝐴−tan²⁡⁡𝐴)
= (sec²⁡⁡𝐴+tan²⁡⁡𝐴−2⁢sec⁡𝐴⁢tan⁡𝐴)/1
= 1 + tan² A + tan² A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan² A = R.H.S

Que-21: (sec A + cosec A)²⁡ + (cos A + sec A)²⁡ = 7 + tan²⁡A + cot²⁡A

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 × (1/cos⁡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-22: sec²⁡⁡A . cosec²⁡⁡A = tan²⁡⁡A + cot²⁡⁡A + 2

Sol: L.H.S. = sec² A . cosec² A
= (1/cos²⁡⁡⁢⁡𝐴) ⋅ (1/sin²⁡⁡⁢⁡𝐴)
= 1/(cos²⁡⁡⁢⁡𝐴.⁢sin²⁡⁡⁢𝐴)
= (sin²⁡⁡⁢⁡𝐴+cos²⁡⁡⁢⁡𝐴)/(cos²⁡⁡⁢⁡𝐴.⁢sin²⁡⁡⁢⁡𝐴)
= (1/cos²⁡⁡⁢⁡𝐴) + (1/sin²⁡⁡⁢⁡𝐴)
= sec² A + cosec² A
= 1 + tan² A + 1 + cot² A   …(∵ sec² A = 1 + tan² A and cosec² A = 1 + cot² A)
= tan² A + cot² A + 2 = R.H.S.

Que-23: {1/(1+cosA)} + {1/(1-cosA)} = 2 cosec²⁡⁡⁢A

Sol: L.H.S. = {1/(1+cos⁡𝐴)} + {1/(1−cos⁡𝐴)}
= (1−cos⁡𝐴+1+cos⁡𝐴)/{(1+cos⁡𝐴)⁢(1−cos⁡𝐴)}
= 2/(1−cos²⁡⁡⁢⁡⁡𝐴)   …(∵ 1 – cos² A = sin² A)
= 2/sin²⁡⁡⁢⁡⁡𝐴
= 2 cosec² A = R.H.S.

Que-24: {1/(1-sinA)} + {1/(1+sinA)} = 2 sec²⁡⁡⁢⁡A

Sol: We have to prove {1/(1+sin⁡𝐴)} + {1/(1−sin⁡𝐴)} = ⁢2 ⁢sec²⁡⁡⁢⁡⁡⁡𝐴
We know that, sin²⁡⁡⁢⁡⁡⁡𝐴 + cos²⁡⁡⁢⁡⁡⁡𝐴 = 1
So,
{1/(1+sin⁡𝐴)} + {1/(1−sin⁡𝐴)} = {(1−sin⁡𝐴)+(1+sin⁡𝐴)}/{(1+sin⁡𝐴)⁢(1−sin⁡𝐴)}
= (1−sin⁡𝐴+1+sin⁡𝐴)/(1−sin²⁡⁡⁢⁡⁡⁡𝐴)
= 2/cos²⁡⁡⁢⁡⁡𝐴
= 2 ⁢sec²⁡⁡⁢⁡⁡⁡𝐴

Que-25: {cosecA/(cosecA-1)} + {cosecA/(cosecA+1)} = 2 sec²⁡⁡⁢⁡⁡A

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⁢𝑐²⁡⁡⁢⁡⁡⁡⁢𝐴+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⁢𝑐²⁡⁡⁢⁡⁡⁡⁢𝐴−1}
= 2⁢cos⁡𝑒⁢𝑐²⁡⁡⁢⁡⁡⁡𝐴/cot²⁡⁡⁢⁡⁡⁡⁡𝐴    …(∵ cosec² A – 1 = cot² A)
= 2⁢ {(1/sin²⁡⁡⁢⁡⁡⁡𝐴)/(cos²⁡⁡⁢⁡⁡⁡⁡𝐴/sin²⁡⁡⁢⁡⁡⁡⁡𝐴)}
= 2/cos²⁡⁡⁢⁡⁡⁡𝐴
= 2 sec² A
= R.H.S.

Que-26: {secA/(secA+1)} + {secA/(secA-1)} = 2 cosec²⁡⁡⁢⁡⁡⁡A

Sol: L.H.S. = {sec⁡𝐴/(sec⁡𝐴+1)} + {sec⁡𝐴/(sec⁡𝐴−1)}
= (sec²⁡⁡⁢⁡⁡⁡⁢𝐴−sec⁡𝐴+sec²⁡⁡⁢⁡⁡⁡⁢⁡𝐴+sec⁡𝐴)/(sec²⁡⁡⁢⁡⁡⁡⁢⁡𝐴−1)
= (2⁢ sec²⁡⁡⁢⁡⁡⁡⁢⁡𝐴)/(tan²⁡⁡⁢⁡⁡⁡⁢𝐴)   …(∵ sec² A – 1 = tan² A)
= (2/cos²⁡⁡⁢⁡⁡⁡⁢⁡𝐴)/(sin²⁡⁡⁢⁡⁡⁡⁢⁡𝐴/cos²⁡⁡⁢⁡⁡⁡⁢⁡𝐴)
= 2/sin²⁡⁡⁢⁡⁡⁡⁢⁡𝐴
= 2 cosec² A = R.H.S.

Que-27: {(1+cosA)/(1-cosA)} = {tan²⁡⁡⁢⁡⁡⁡⁢A/(secA-1)²⁡⁡⁢⁡⁡⁡⁢}

Sol: R.H.S. = {tan²⁡⁡⁢⁡⁡⁡⁢⁡𝐴/(sec⁡𝐴−1)²⁡⁡⁢⁡⁡⁡⁢⁡}
= {(sec²⁡⁡⁢⁡⁡⁡⁢⁡𝐴−1)/(sec⁡𝐴−1)²⁡⁡⁢⁡⁡⁡⁢⁡}    …[sec²θ – tan²θ = 1 sec²θ – 1 = tan²θ]
= {(sec⁡𝐴+1)⁢(sec⁡𝐴−1)}/{(sec⁡𝐴−1)²⁡⁡⁢⁡⁡⁡⁢⁡}
= (sec⁡𝐴+1)/(sec⁡𝐴−1)
= {(1/cos⁡𝐴)+1} / {(1/cos⁡𝐴)−1}
= {(1+cos⁡𝐴)/cos⁡𝐴} / {(1−cos⁡𝐴)/cos⁡𝐴}
= (1+cos⁡𝐴)/(1−cos⁡𝐴)
R.H.S. = L.H.S.

Que-28: cot²⁡⁡⁢⁡⁡⁡⁢⁡A/(cosecA+1)²⁡⁡⁢⁡⁡⁡⁢⁡ = (1-sinA)/(1+sinA)

Sol: R.H.S. = (1−sin⁡𝐴)/(1+sin⁡𝐴)
= {1−(1/cos⁡𝑒⁢𝑐⁢𝐴)} / {1+(1/cos⁡𝑒⁢𝑐⁢𝐴)}
= (cos⁡𝑒⁢𝑐⁢𝐴−1)/(cos⁡𝑒⁢𝑐⁢𝐴+1)
= (cos⁡𝑒⁢𝑐⁢𝐴−1)/(cos⁡𝑒⁢𝑐⁢𝐴+1) × (cos⁡𝑒⁢𝑐⁢𝐴+1)/(cos⁡𝑒⁢𝑐⁢𝐴+1)
= (cos⁡𝑒⁢𝑐²⁡⁡⁢⁡⁡⁡⁢⁡⁢𝐴−1)/{(cos⁡𝑒⁢𝑐⁢𝐴+1)²⁡⁡⁢⁡⁡⁡⁢⁡}
= cot²⁡⁡⁢⁡⁡⁡⁢⁡⁡𝐴/(cos⁡𝑒⁢𝑐⁢𝐴+1)²⁡⁡⁢⁡⁡⁡⁢⁡  …(∵ cosec² A – 1 = cot² A)
= L.H.S.

Que-29: {(1+sinA)/cosA} + {cosA/(1+sinA)} = 2 secA

Sol: L.H.S. = {(1+sin⁡𝐴)/cos⁡𝐴} + {cos⁡𝐴/(1+sin⁡𝐴)}
= {(1+sin⁡𝐴)²⁡⁡⁢⁡⁡⁡⁢⁡⁢+cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢𝐴}/{cos⁡𝐴⁢(1+sin⁡𝐴)}
= {1+sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡𝐴+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-30: (1-sinA)/(1+sinA) = (sec A – tan A)²⁡⁡⁢⁡⁡⁡⁢⁡⁢

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}²⁡⁡⁢⁡⁡⁡⁢⁡⁢
= {(1/cos⁡𝐴)−(sin⁡𝐴/cos⁡𝐴)}²⁡⁡⁢⁡⁡⁡⁢⁡⁢
= (sec 𝐴 − tan ⁡𝐴)²⁡⁡⁢⁡⁡⁡⁢⁡⁢
= R.H.S.

Que-31: (cot A – cosec A)²⁡⁡⁢⁡⁡⁡⁢⁡⁢ = (1-cosA)/(1+cosA)

Sol: (cot⁡𝐴−cos⁡𝑒⁢𝑐⁢𝐴)²⁡⁡⁢⁡⁡⁡⁢⁡⁢ = (1−cos⁡𝐴)/(1+cos⁡𝐴)
cot²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡𝐴 − 2⁢cot⁡𝐴 ⁢cos⁡𝑒⁢𝑐⁢𝐴 + cos⁡𝑒⁢𝑐²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁢𝑥 = (1−cos⁡𝐴)/(1+cos⁡𝐴)
(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢𝐴/sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡𝐴) − (2⁢cos⁡𝐴/sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡𝐴) + (1/sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢/⁡𝐴) = (1−cos⁡𝐴)/(1+cos⁡𝐴)
(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢𝐴−2⁢cos⁡𝐴+1)/sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡𝐴 = (1−cos⁡𝐴)/(1+cos⁡𝐴)
(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡𝐴−2⁢cos⁡𝐴+1)/(1−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢𝐴) = (1−cos⁡𝐴)/(1+cos⁡𝐴)
{(1−cos⁡𝐴)⁢(1−cos⁡𝐴)} / {(1+cos⁡𝐴)⁢(1−cos⁡𝐴)} = (1−cos⁡𝐴)/(1+cos⁡𝐴)
(1−cos⁡𝐴)/(1+cos⁡𝐴) = (1−cos⁡𝐴)/(1+cos⁡𝐴)

Que-32: (cosec A – 1)/(cosec A + 1) = {cos A/(1+sinA)}²

Sol: L.H.S. = (cos⁡𝑒⁢𝑐⁢𝐴−1)/(cos⁡𝑒⁢𝑐⁢𝐴+1)
= {(cos⁡𝑒⁢𝑐⁢𝐴−1)/(cos⁡𝑒⁢𝑐⁢𝐴+1)} × {(cos⁡𝑒⁢𝑐⁢𝐴+1)/(cos⁡𝑒⁢𝑐⁢𝐴+1)}
= (cos⁡𝑒⁢𝑐²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴−1)/{(cos⁡𝑒⁢𝑐⁢𝐴+1)²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡}
= cot²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁡ 𝐴 / (cos⁡𝑒⁢𝑐⁢𝐴+1)²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡
= (cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁡𝐴/sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁡𝐴) / {((1/sin⁡𝐴)+1)²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡}
= {cos⁡𝐴/(1+sin⁡𝐴)}²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡ = R.HS.

Que-33: tan²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡ A – tan²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡ B = (sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡A-sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡B)/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡A.cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡B)

Sol: LHS = tan²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴 − tan²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵
= (sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴/cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴) − (sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵/cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
= (sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴⁢sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
= {(1−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴)⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴⁢(1−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)}/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐵)
= {cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴+cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡B}/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
= (cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴)/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
= {(1−sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)−(1−sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴)} / (cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
= (sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴−sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐵)/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
Hence tan²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴 − tan²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵 = (cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢𝐵−cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴)/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)
= (sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴−sin²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)/(cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐴.⁢cos²⁡⁡⁢⁡⁡⁡⁢⁡⁢⁡⁢⁡𝐵)

Que-34: (sinθ – 2sin³θ)/(2cos³θ – cosθ) = tanθ

Sol: LHS = (sin⁡θ−2⁢sin³θ) / (2⁢cos³⁡θ−cos⁡θ)
= {sin⁡θ⁢(1−2⁢sin²θ)} / {cos⁡θ⁢(2⁢cos²⁡θ−1)}
= {tan⁡θ⁢(1−2⁢(1−cos²⁡θ))} / (2⁢cos²⁡θ−1)
= {tan⁡θ⁢(1−2+2⁢cos²⁡θ)} / (2⁢cos²θ−1)
= {tan⁡θ⁢(2⁢cos²⁡θ−1)} / (2⁢cos²⁡θ−1)
= tan θ
= RHS
Hence proved.

Que-35: sin A / (1 + cos A) = cosec A – cot A

Sol: L.H.S. = sin⁡𝐴/(1+cos⁡𝐴)
= {sin⁡𝐴/(1+cos⁡𝐴)} × {(1−cos⁡𝐴)/(1−cos⁡𝐴)}
= {sin⁡𝐴⁢(1−cos⁡𝐴)}/{1−cos²⁡𝐴}
= {sin⁡𝐴⁢(1−cos⁡𝐴)}/sin²⁡𝐴
= (1−cos⁡𝐴)/sin⁡𝐴
= (1/sin⁡𝐴) − (cos⁡𝐴/sin⁡𝐴)
= cosec A – cot A = R.H.S.

Que-36: cos A / (1 – sin A) = sec A + tan A

Sol: RHS = sec A + tan A
= (1/cosA) + (sinA/cosA) = (1+sinA)/cosA
= {= (1+sinA)/cosA} × {(1-sinA)/(1-sinA)}
= {1-sin²A}/{cosA/(1-sinA)}
= cos²A / {cosA/(1-sinA)}
= cos A / (1 – sin A)
= LHS

Que-37: (sin A tan A)/(1 – cos A) = 1 + sec A

Sol:  L.H.S. = (sin⁡𝐴 ⁢tan⁡𝐴)/(1−cos⁡𝐴)
= {(sin⁡𝐴tan⁡𝐴)/(1−cos⁡𝐴)} × {(1+cos⁡𝐴)/(1+cos⁡𝐴)}
= {sin⁡𝐴⁢ tan⁡𝐴⁢(1+cos⁡𝐴)} / {1−cos²⁡𝐴}
= {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-38: (1 + cot A – cosec A) (1 + tan A + sec A) = 2

Sol: L.H.S. = (1 + cot A – cosec A)(1 + tan A + sec A)
= {1 + (cos⁡𝐴/sin⁡𝐴) − (1/sin⁡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−1)⁢(sin⁡𝐴+cos⁡𝐴+1)} / (sin⁡𝐴.⁢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⁢ sin⁡𝐴 ⁢cos⁡𝐴) / (sin⁡𝐴 ⁢cos⁡𝐴)
= 2 = R.H.S

Que-39: √{(1+sinA)/(1-sinA)} = sec A + tan A

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−sin²𝐴)}
= √{(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-40: √{(1-cosA)/(1+cosA)} = cosec A – cot A

Sol: L.H.S. = √{(1−cos⁡𝐴)/(1+cos⁡𝐴)}
= √[{(1−cos⁡𝐴)/(1+cos⁡𝐴)} × {(1−cos⁡𝐴)/(1−cos⁡𝐴)}]
= √{(1−cos⁡𝐴)²/(1−cos²⁡𝐴)}
= √{(1−cos⁡𝐴)²/sin²⁡𝐴}
= (1−cos⁡𝐴)/sin⁡𝐴
= (1/sin⁡𝐴) − (cos⁡𝐴/sin⁡𝐴)
= cosec A – cot A = R.H.S.

Que-41: √{(1-cosA)/(1+cosA)} = sin A / (1 + cos A)

Sol: 𝐿.𝐻.𝑆 = √{(1−cos⁡𝐴)/(1+cos⁡𝐴)}
= √[{(1−cos⁡𝐴)⁢(1+cos⁡𝐴)}/{(1+cos⁡𝐴)⁢(1+cos⁡𝐴)}]
=√{(1−cos²⁡𝐴)/(1+cos⁡𝐴)²}
= √{sin²𝐴/(1+cos⁡𝐴)²
= sin⁡ 𝐴 / 1 + cos⁡ 𝐴
= R.H.S

Que-42: √{(1-sinA)/(1+sinA)} = cos A / (1 + sin A)

Sol: L.H.S. = √{(1−sin⁡𝐴)/(1+sin⁡𝐴)}
= √[{(1−sin⁡𝐴)/(1+sin⁡𝐴)} × {(1+sin⁡𝐴)/(1+sin⁡𝐴)}]
= √{(1−sin²⁡𝐴)/(1+sin⁡𝐴)²}
= √{cos²⁡𝐴/(1+sin⁡𝐴)²
= cos ⁡𝐴 / (1 + sin ⁡𝐴) = R.H.S.

Que-43: 1 – {cos²A/(1+sinA)} = sinA

Sol: L.H.S. = 1 − {cos²⁡⁡𝐴/(1+sin⁡𝐴)}
= {1+sin⁡𝐴−cos²⁡⁡𝐴}/(1+sin⁡𝐴)
= (sin⁡𝐴+sin²⁡⁡𝐴)/(1+sin⁡𝐴)
= {sin⁡𝐴⁢(1+sin⁡𝐴)}/(1+sin⁡𝐴)
= sin A = R.H.S.

Que-44: {1/(sinA+cosA)} + {1/(sinA-cosA)} = {2sinA/(1-2cos²A)}

Sol: L.H.S. = {1/(sin⁡𝐴+cos⁡𝐴)} + {1/(sin⁡𝐴−cos⁡𝐴)}
= (sin⁡𝐴−cos⁡𝐴+sin⁡𝐴+cos⁡𝐴) / {(sin⁡𝐴+cos⁡𝐴)⁢(sin⁡𝐴−cos⁡𝐴)}
= 2⁢ sin⁡ 𝐴 / (sin²⁡⁡𝐴−cos²⁡⁡𝐴)
= 2 ⁢sin ⁡𝐴 / {1−cos²⁡⁡𝐴−cos²⁡⁡⁡𝐴}     …(∵ sin²A = 1 – cos²A)
= 2 ⁢sin ⁡𝐴 / (1−2⁢cos²⁡⁡𝐴)
= R.H.S.

Que-45: {(sinA+cosA)/(sinA-cosA)} + {(sinA-cosA)/(sinA+cosA)} = 2 / (2sin²⁡ ⁡A – 1)

Sol: LHS = {(sin⁡𝐴+cos⁡𝐴)/(sin⁡𝐴−cos⁡𝐴)} + {(sin⁡𝐴−cos⁡𝐴)/(sin⁡𝐴+cos⁡𝐴)}
= {(sin⁡𝐴+cos⁡𝐴)²⁡⁡+(sin⁡𝐴−cos⁡𝐴)²⁡⁡} / {(sin⁡𝐴−cos⁡𝐴)⁢(sin⁡𝐴+cos⁡𝐴)}
=  {sin²⁡⁡⁡𝐴+cos²⁡⁡⁡𝐴+2⁢sin⁡𝐴⁢cos⁡𝐴+sin²⁡⁡𝐴+cos²⁡⁡⁡𝐴−2⁢sin⁡𝐴.cos⁡𝐴} / (sin²⁡⁡⁡𝐴−cos²⁡⁡⁡𝐴)
= {2⁢(sin²⁡⁡⁡𝐴+cos²⁡⁡𝐴)} / (sin²⁡⁡⁡𝐴−cos²⁡⁡⁡𝐴)
= (2×1) / {sin²⁡⁡⁡𝐴−(1−sin²⁡⁡⁡𝐴)}
= 2 / (sin²⁡⁡⁡𝐴−1+sin²⁡⁡⁡𝐴)
= 2 / (2⁢sin²⁡⁡⁡𝐴−1)
= RHS
Hence proved.

Que-46: (cotA+cosecA-1) / (cotA-cosecA+1) = (1+cosA)/sinA

Sol: L.H.S. = (cot⁡𝐴+cos⁡𝑒⁢𝑐⁢𝐴−1) / (cot⁡𝐴−cos⁡𝑒⁢𝑐⁢𝐴+1)
= {cot⁡𝐴+cos⁡𝑒⁢𝑐⁢𝐴−(cos⁡𝑒⁢𝑐²⁢𝐴−cot²⁡𝐴)} / {cot⁡𝐴−cos⁡𝑒⁢𝑐⁢𝐴+1}  …[cosec²A – cot²A = 1]
= {cot⁡𝐴+cos⁡𝑒⁢𝑐⁢𝐴−[(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)⁢(cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴)]} / (cot⁡𝐴−cos⁡𝑒⁢𝑐⁢𝐴+1)
= {cot⁡𝐴+cos⁡𝑒⁢𝑐⁢𝐴⁢(1−cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴)} / (cot⁡𝐴−cos⁡𝑒⁢𝑐⁢𝐴+1)
= cot A + cosec A
= (cos⁡𝐴/sin⁡𝐴) + (1/sin⁡𝐴)
= (1+cos⁡𝐴)/sin⁡𝐴

Que-47: {(1+sinA)/(cosecA-cotA)} = {(1-sinA)/(cosecA+cotA)} = 2(1+cotA)

Sol: L.H.S. = {(1+sin⁡𝐴)/(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴) − {(1−sin⁡𝐴)/(cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴)}
= {(1+sin⁡𝐴)⁢(cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴)−(1−sin⁡𝐴)⁢(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)} / {(cos⁡𝑒⁢𝑐⁢𝐴−cot⁡𝐴)⁢(cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴)}
= (cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴+sin⁡𝐴⁢cos⁡𝑒⁢𝑐⁢𝐴+sin⁡𝐴⁢cot⁡𝐴−cos⁡𝑒⁢𝑐⁢𝐴+cot⁡𝐴+sin⁡𝐴⁢cos⁡𝑒⁢𝑐⁢𝐴−sin⁡𝐴⁢cos⁡𝐴) / (cos⁡𝑒⁢𝑐⁢²⁢𝐴−cot²⁢⁡𝐴)
= 2 cot A + 2 sin A cosec A
= 2 cot A + 2 1cos⁡𝑒⁢𝑐⁢𝐴 ×cos⁡𝑒⁢𝑐⁢𝐴
= 2 (cot A + 1)
Hence proved.

Que-48: (cos θ cot θ)/(1+sin θ) = cosec θ – 1

Sol: L.H.S. = (cos ⁡𝜃 ⁢cot ⁡𝜃)/(1+sin ⁡𝜃)
= {(cos⁡𝜃 ⁢cot⁡𝜃)/(1+sin⁡ 𝜃)} × {(1−sin⁡𝜃)/(1−sin⁡𝜃)}
= {cos⁡𝜃⁢ cot⁡𝜃⁢(1−sin⁡𝜃)}/(1−sin²⁢⁡⁡𝜃)
= {cos⁡𝜃 (⁢cos⁡𝜃/sin⁡𝜃)⁢ (1−sin⁡𝜃)} / cos²⁢⁡𝜃
= (1−sin⁡𝜃) / sin⁡𝜃
= (1/sin ⁡𝜃) − 1
= cosec θ – 1

–: Trigonometrical Identities Class 10 Concise Exe-21A ICSE Maths Selina Solutions :–

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