Chapter 21 Trigonometric Identities Questions and Answers Class 10 Maths

ICSE Solutions for Chapter 21 Trigonometric Identities Class 10 Mathematics

Question 1: Use tables to find sine of: 
(i) 21° 
(ii) 34°42 
(iii) 47° 32' 
(iv) 62° 57' 
(v) 10° 20' + 20° 45'

Solution 1:
(i) sin 21° = 0.3584
(ii) sin 34° 42' = 0.5693
(iii) sin 47° 32' = sin (47° 30' + 2')  = 0.7373 + 0.0004 = 0.7377
(iv) sin 62°57' = sin (62° 54' + 3') = 0.8902 + 0.0004 = 0.8906
(v) sin (10° 20' + 20° 45') = sin 30°65' = sin 31˚5' = 0.5150 + 0.0012 = 0.5162

Question 2:  Express each of the following in terms of angles between 0 and 45°: 
(i) sin 59° + tan 63° 
(ii) cosec 68° + cot 72° 
(iii) cos 74° +  sec 67°

Solution 2:
(i) sin 59˚ + tan 63˚
= sin(90 - 31)˚ + tan(90 – 27)˚
=  cos 31° + cot 27°

(ii) cosec 68° + cot 72°
= cosec (90 – 22)˚+ cot(90 - 18)˚
= sec 22° + tan 18°

(iii) cos 74° + sec 67°
=  cos(90 - 16)˚ + sec(90 - 23)˚
= sin 16° + cosec 23°

Question 3: If m = a secA + b tanA and n = a tanA + b secA, prove that m² - n² = a² - b²  

Solution 3: Given,
m = a secA + b tanA and n = a tanA + b secA
m² – n²  = (a secA + b tanA)² - (a tanA + b secA)² 
= a² sec²A + b²  tan² A + 2ab secAtanA
= (a² tan² A + b²sec² A + 2ab secA tanA)
= sec²A(a² - b²) + tan²A(b² – a²)
= (a² – b²)[sec²A – tan²A]
= (a² – b²) [Since sec²A – tana²A = 1]
Hence, m² - n² = a² - b²

Question 4: Prove:
tanA - cotA = (1 - 2 cos²A)/(SinACos A)

Solution 4: tanA – cotA = (sinA cosA – cosA/sinA)
= (sin²A – cos²A)/(sinAcosA)
= (1 – cos²A – cos²A)/(sinAcosA)  (∵  sin²A = 1 – cos²A)
= (1 - 2cosb²A)/ sinAcosA

Question 5: Prove: sin⁴A – cos⁴A = 2sin²A - 1 

Solution 5: sin⁴A – cos⁴A
= (sin²A)² - (cos²A)²
= (sin²A + cos²A) (sin²A – cos²A)
= sin²A – cos²A
= sin²A - (1 – sin²A)
= 2 sin²A - 1

Question 6: If x = r cosA cosB, y = r cosA sinB and z = r sin A, prove that x² + y² + z² = r²

Solution 6: 
LHS = (rcosAcosB)² + (rcosAsinB)² + (rsinA)²
= r²cos² A cos² B + r²cos² A sin² B + r²sin² A
= r²cos² A(cos²B + sin²B) + r² sinb² A
= r²(cos²A + sin²A) = r² = RHS

Question 7: 1f cosA/cosB = m and cosA/cosB  = n, show that (m² + n²) cos²B = n² 

Solution 7: LHS = (m² + n²) cos²B
= (cos²A/cos²B + cos²A/sin²B).(cos²B)
= (cos²A sin²B + cos²A cos²B)/(cos²Bsin²B).cos²B
= (cos²A sin²B + cos²A cos² B)/(sin²B)
= cos²A(sin²B + cos²B)/(sin²B)
= cos²A/sin²B
= n²
Hence, (m² + n²) cos²B = n²

Question 8: In each case, given below, find the value of angle A, where 0° ≤ A ≤ 90° 
(i) sin(90° - 3A).cosec42˚ = 1 
(ii) cos(90° - A). sec77˚ = 1

Solution 8: (i) sin(90° - 3A).cosec42˚ = 1
(cos 3A). 1/(sin42˚) = 1
cos 3A = sin42˚ = sin(90˚ – 48˚) = cos48˚
3A = 48˚
A = 16˚

(ii) cos(90˚ – A).sec77˚ = 1
cos(90˚ - A).sec77˚ = 1
sinA.(1/cos77˚) = 1
sinA = cos77˚ = cos(90˚ – 13˚) = sin13˚
A = 13˚

Question 9: Prove that:  
(i) {cos(90° - θ)cos θ}/cotθ = 1 – cos²θ
(ii) sinθ sin(90° - θ)/cot(90˚ - θ) = 1 – sin²θ

Solution 9:
(i) L.H.S = {cos(90˚ - θ)cosθ}/cotθ = (sinθcosθ)/(cosθ/sinθ) = sin²θ = 1 – cos²θ
(ii) L.H.S = sinθsin(90˚ - θ)/cot(90˚ - θ) = (sinθcosθ)/tanθ = (sinθcosθ)/(sinθ/cosθ) = cos²θ = 1 – sin²θ

Question 10: Prove that:
(i) 1/(sinA – cosA) – 1/(sinA + cosA) = 2cosA/(2sin²A – 1)
(ii) cot²A/(cosecA – 1) – 1 = cosecA
(iii) cosA/(1 + sinA) = secA – tanA 
(iv) cosA(1 + cotA) + sinA(1 + tanA) = secA + cosecA
(v) (sin A – cosA)(1 + tanA + cot A) = secA/cosec²A – cosecA/sec²A
(vi) 

(vii) (sinA + cosA)(secA+ cosecA) = 2 + secA cosecA

(vii)(tan A+ cotA)(cosecA - sin A)(sec A - cosA) = 1
(ix) cot²A – cot²B = (cos²A – cos²B)/(sin²Asin²B) = (cosec²A – cosec²B) 

Solution 10:
(i) 1/(sinA – cosA) – 1/(sinA + cosA)
= (sinA + cosA – sinA + cosA)/(sinA – cosA)(sinA + cosA)
= 2cosA/(sin²A – cos²A)
= 2cosA/(sin²A – (1 – sin²A)
= 2cosA/(2sin²A – 1)

(ii) cot²A/(cosecA – 1) = 1
= (cot²A – cosecA + 1)/(cosecA – 1)
= (-cosecA + cosec²A)/(cosecA – 1)
= cosecA(cosecA – 1)/(cosecA – 1)
= cosecA

(iii) cosA/(1 + sinA)
= cosA/(1 + sinA) × (1 – sinA)/(1 – sinA)
= cosA(1 – sinA)(1 – sin²A)
= cosA(1  - sinA)/(cos²A)
= (1 – sinA)/cosA
= secA – tanA

(iv) cosA(1 + cotA) + sinA(1 + tanA)
= cosA + cos²A/sinA + sinA + sin²A/cosA
= sinA + cos²A/sinA + cosA + sin²A/cosA
= {(cos²A + sin²A)/sinA} + {(cos²A + sin²A)/cosA}
= 1/sinA + 1/cosA
= cosec A + secA

(v) (sinA – cosA)(1 + tanA + cotA)
= sinA + sin²A/cosA + cosA – cosA – sinA – cos²A/sinA
= (sin²A/cosA – cos²A/sinA)
= secA/cosec²A – cosecA/sec²A

R.H.S. = tanA + cotA
= (sinA/cosA + cosA/sinA)
= (sin²A + cos²A)/(sinAcosA)
= 1/(sinAcosA)
L.H.S = R.H.S

(vii) (sinA + cosA)(secA + cosecA)
= sinA/cosA + 1 + 1 + cosA/sinA
= {2 + (cos²A + sin²A)}/(sinAcosA)
= 2 + 1/(sinAcosA)
= 2 + secAcosecA

(viii) (tanA + cotA)(cosecA – sinA)(secA – cosA)
= (sinA/cosA + cosA/sinA)(1/sinA – sinA)(1/cosA – cosA)
= {(sin²A + cos²A)/(sinAcosA)}{(1–sin²A)/(sinA)}{(1–cos²A)/(cosA)}
= (1/sinAcosA)(cos²A/sinA)(sin²A/cosA)
= 1

(ix) cot²A – cot²B
= (cos²A/sin²A – cos²B/sin²B)
= (cos²A sin²B – cos²B sin²A)/(sin²Asin²B)
= (cos²A(1 – cos²B) – cos²B(1 – cos²A)/(sin²Asin²B)
= (cos²A – cos²A cos²B – cos²B + cos²Bcos²A)/(sin²Asin²B)
= (cos²A – cos²B)/(sin²A sin²B)
= (1 – sin²A – 1 + sin²B)/(sin²A sin²B)
= (- sin²A + sin²B)/(sin²A sin²B)
= (sin²B/sin²A sin²B) – (sin²A/sin²A sin²B)
= 1/(sin²A) – 1/(sin²B)
= cosec²A – cosec²B