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 m- n2 = a- b

Solution 3: Given,
m = a secA + b tanA and n = a tanA + b secA
m– n = (a secA + b tanA)2 - (a tanA + b secA)
= asec2A + b tanA + 2ab secAtanA
= (atan2 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° - Î¸)cos Î¸}/cotÎ¸ = 1 – cos2Î¸
(ii) sinÎ¸ sin(90° - Î¸)/cot(90˚ - Î¸) = 1 – sin2Î¸

Solution 9:
(i) L.H.S = {cos(90˚ - Î¸)cosÎ¸}/cotÎ¸ = (sinÎ¸cosÎ¸)/(cosÎ¸/sinÎ¸) = sin2Î¸ = 1 – cos2Î¸
(ii) L.H.S = sinÎ¸sin(90˚ - Î¸)/cot(90˚ - Î¸) = (sinÎ¸cosÎ¸)/tanÎ¸ = (sinÎ¸cosÎ¸)/(sinÎ¸/cosÎ¸) = cos2Î¸ = 1 – sin2Î¸

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