Question 1.
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-465.png)
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-471.png)
Question 2.
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-467.png)
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-470.png)
Question 3.
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-472.png)
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-473.png)
Question 4.
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-474.png)
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-475.png)
Question 5. Prove: sin4 A – cos2 A = 2sin2 A – 1
Solution:
sin4 A – cos4 A
(sin2 A)2-(cos2 A)2
(sin2 A + cos2 A) – (sin2 A – cos2 A)
1-(sin2 A-cos2 A)
sin2 A – cos2 A
sin2 A – (1- sin2 A)
2sin2 A – 1
LHS = RHS
Question 6. Prove: (1 – tanA)2 + (1 + tanA)2 = 2sec2 A
Solution:
From LHS,
(1-tanA)2 + (1 + tanA)2
We know that, (a-b)2 = a2 + b2 – 2ab
(1 + tan2 A-2tanA) + (1 + tan2 A + 2tanA)
2(1 + tan2 A)
2sec2 A = RHS
Question 7. Prove: cosec4 A-cosec2 A=cot4 A + cot2 A
Solution:
From LHS,
cosec2 A(cosec2 A – 1)
From RHS,
cot4 A + cot2 A
cot2 A(cot2 A + 1)
(cosec2 A – 1)cosec2 A
LHS = RHS
Question 8. Prove: secA(1 – sinA)(secA + tanA) = 1
Solution
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-476.png)
Question 9. Prove: cosecA(1 + cosA)(cosecA – cotA) = 1
Solution:
From LHS
cosecA(1 + cosA)(cosecA – cotA) = 1
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-477.png)
Question 10. Prove: sec2 A + cosec2 A=sec2 Acosec2 A
Solution:
From LHS
sec2 A + cosec2 A
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-478.png)
Question 11.
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-479.png)
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-480.png)
Question 12. Prove: tan2 A – sin2 A = tan2 Asin2 A
Solution:
From LHS,
tan2 A – sin2 A
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-481.png)
Question 13. Prove: cot2A − cos2A = cos2A cot2A
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-482.png)
Question 14. Prove: (cosecA + sinA)(cosecA − sinA) = cot2A + cos2A
Solution:
From LHS
(cosecA + sinA)(cosecA − sinA)
cosec2A − sin2A
(1 + cot2A) − (1 − cos2A)
1 + cot2A − 1 + cos2A
cot2A + cos2A
LHS = RHS
Question 15. Prove: (secA − cosA)(secA + cosA) = sin2A + tan2A
Solution:
From LHS
(secA − cosA)(secA + cosA)
We know that, (a + b)(a − b) = a2 − b2
sec2A − cos2A
(1 + tan2A) − (1 − sin2A)
sin2A − tan2A
LHS = RHS
Question 16. Prove: (cosA + sinA)2 + (cosA − sinA)2 = 2
Solution:
From LHS
(cosA + sinA)2 + (cosA − sinA)2
cos2A + sin2A + 2cosA sinA + cos2A + sis2A − 2cosA sinA
cos2A + sin2A + cos2A + sin2A
2cos2A + 2sin2A
2(cos2A + sin2A)
2 × 1 = 2
LHS = RHS
Question 17. Prove: (cosecA − sinA)(secA − cosA)(tanA + cotA) = 1
Solution:
From LHS
(cosecA − sinA)(secA − cosA)(tanA + cotA)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-483.png)
Question 18.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-484.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-485.png)
Question 19.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-486.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-487.png)
Question 20.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-488.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-489.png)
Question 21. Prove: (sinA + cosecA)2 + (cosA + secA)2 = 7 + tan2A + cot2A
Solution:
From LHS
(sinA + cosecA)2 + (cosA + secA)2
sin2A + cosec2A + 2sinAcosecA + cos2A + sec2A + 2cosAsecA
sin2A + cosec2A + 2 + cos2A + sec2A + 2
sin2A + cosec2A + cos2A + sec2A + 4
sin2A + cos2A + cosec2A + sec2A + 4
1 + cosec2A + sec2A + 4
cosec2A + sec2A + 5
1 + cot2A + 1 + tan2A + 5
cot2A + 1 + tan2A + 6
7 + tan2A + cot2A
LHS = RHS
Question 22. Prove: sec2A. cosec2A = tan2A + cot2A + 2
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-490.png)
Question 23.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-491.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-492.png)
Question 24.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-493.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-494.png)
Question 25.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-495.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-496.png)
Question 26.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-497.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-498.png)
Question 27.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-499.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-500.png)
Question 28.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-502.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-503.png)
Question 29.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-504.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-505.png)
Question 30.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-506.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-507.png)
Exercise 21 B
Question 1. Prove:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-508.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-509.png)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-510.png)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-511.png)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-512.png)
(v) 2sin2A + cos4A
2sin2A + (1 − sin2A)2
2sin2A + 1 + sin4A − 2sin2A
1 + sin4A
LHS = RHS
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-513.png)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-514.png)
(viii) (1 + tanAtanB)2 + (tanA − tanB)2
1 + tan2A tan2B + 2tanAtanB + tan2A + tan2B − 2tanAtanB
1 + tan2A tan2B + tan2 A + tan2B
sec2A + tan2B(1 + tan2A)
sec2A + tan2B sec2A
sec2A(1 + tan2B)
sec2A sec2B
LHS = RHS
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-515.png)
Question 2. If xcosA + ysinA = m and xsinA − ycosA = n, then prove that x2 + y2 = m2 + n2.
Solution:
m2 + n2
(xcosA + ysinA)2 + (xsinA − ycosA)2
x2cos2A + y2sin2A + 2xysinAcosA + x2sin2A + y2cos2A − 2xysinAcosA
x2cos2A + x2sin2A + y2sin2A + y2cos2A
x2(cos2A + sin2A) + y2(cos2A + sin2A)
x2 + y2
Therefore, x2 + y2 = m2 + n2
Question 3. If m = asecA + btanA and n = atanA + bsecA, prove that m2 − n2 = a2 − b2
Solution:
It is given that,
m = asecA + btanA and n = atanA + bsecA
m2 − n2 = (asecA + btanA)2 − (atanA + bsecA)2
a2sec2A + b2tan2A + 2ab secAtanA − (a2tan2A + b2sec2A + 2ab secAtanA)
a2sec2A + b2tan2A + 2ab secAtanA − a2tan2A − b2sec2A − 2ab secAtanA
a2sec2A + b2tan2A − a2tan2A − b2sec2A
a2sec2A − b2sec2A + b2tan2A − a2tan2A
sec2A(a2 − b2) + tan2A(a2 − b2)
(a2 − b2){sec2A + tan2A}
(a2 − b2)
Therefore, m2 − n2 = a2 − b2
Question 4. If x = rsinAcosB,y = rsinAsinB and z = r cosA, prove that x2 + y2 + z2 = r2
Solution:
From LHS
(rsinAcosB)2 + (rsinAsinB)2 + (rcosA)2
r2sin2Acos2B + r2sin2Asin2B + r2cos2A
r2sin2A(cos2B + sin2B) + r2cos2A
r2sin2A + r2cos2A
r2(sin2A + cos2A)
r2
LHS = RHS
Question 5. If sinA + cosA = m and secA + cosecA = n, prove that n(m2−1) = 2m.
Solution:
It is given that,
sinA + cosA = m and secA + cosecA = n
From LHS,
n(m2 − 1)
(secA + cosecA)[(sinA + cosA)2−1]
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-516.png)
Question 6. If x = rcosAcosB,y = rcosAsinB and z = rsinA, Prove that x2 + y2 + z2 = r2.
Solution:
From LHS
(rcosAcosB)2 + (rcosAsinB)2 + (rsinA)2
r2cos2Acos2B + r2cos2A sin2B + r2sin2A
r2cos2A (cos2B + sin2B) + r2sin2A
r2cos2A + r2sin2A
r2(cos2A + sin2A)
r2
LHS = RHS
Question 7.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-517.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-518.png)
Question 1. Without using trigonometric tables, show that:
(i) tan10˚ tan15˚ tan75˚ tan80˚ = 1
(ii) sin42˚ sec48˚ + cos42˚cosec48˚ = 2
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-520.png)
Solution:
(i) tan10˚tan15˚tan75˚tan80˚
tan(90˚-80˚)tan(90˚-75˚)tan75˚tan80˚
cot80˚cot75˚tan75˚tan80˚
We know that, cotƟtanƟ = 1
1
LHS=RHS
(ii) sin42˚ sec48˚+cos42˚ cosec48˚ = 2
sin42˚ sec(90˚-42˚) + cos42˚ cosec(90˚-42˚)
sin42˚ cosec42˚ + cos42˚ sec42˚
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-521.png)
Question 2. Express each of the following in terms of angles between 0˚ and 45˚:
(i) sin59˚ + tan63˚
(ii) cosec68˚ + cot72˚
(iii) cos74˚ + sec67˚
Solution:
(i) sin59˚ + tan63˚
sin(90˚ − 31˚) + tan(90˚ − 27˚)
cos31˚ + cot27˚
LHS = RHS
(ii) cosec68˚ + cot72˚
cosec(90˚ − 22˚) + cot(90˚ − 18˚)
sec22˚ + tan18˚
LHS = RHS
(iii) cos74˚ + sec67˚
cos(90˚ − 16˚) + sec(90˚ − 23˚)
sin16˚ + cosec23˚
LHS = RHS
Question 3. Show that:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-522.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-523.png)
Question 4. For triangle ABC, show that:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-524.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-525.png)
Question 5. A triangle ABC is right angled at B; find the value of
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-526.png)
Solution:
It is given that, ABC is a right angled triangle, right angled at B.
A + C = 90˚
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-527.png)
Question 6. (i) sinx = sin60˚cos30˚ – cos60˚sin30˚
(ii) sinx = sin60˚cos30˚ + cos60˚sin30˚
(iii) cosx = cos60˚cos30˚ – sin60˚sin30˚
(iv) tanx = tan60˚ – tan30˚/1 + tan60˚ tan30˚
(v) sin2x = 2sin45˚ + cos45˚
(vi) sin3x = 2sin30˚ + cos30˚
(vii) cos(2x – 6˚) = cos2 30˚ – cos2 60˚
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-528.png)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-529.png)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-530.png)
(vii) cos(2x − 6) = cos230˚ − cos260˚
cos(2x − 6) = cos2(90˚ − 60˚) − cos260˚
cos(2x − 6) = sin260˚ − cos260˚
cos(2x − 6) = 1 − 2cos260˚
cos(2x − 6) = 1 − 2(12)2
cos(2x − 6) = 1 − 2(14)
cos(2x − 6) = 1 − 12
cos(2x − 6) = 12
cos(2x − 6) = cos60˚
2x − 6 = 60˚
2x = 60˚ + 6˚
2x = 66˚
x = 33˚
Question 7. 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:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-531.png)
Question 8. Prove that:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-532.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-533.png)
Question 9.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-534.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-535.png)
Question 10. Evaluate:
sin234˚ + sin256˚ + 2tan18˚ tan72˚ − cot230˚
Solution:
sin234˚ + sin256˚ + 2tan18˚ tan72˚ − cot230˚
sin234˚ + sin2(90˚ − 34˚) + 2tan18˚ tan (90˚ − 72˚) − cot230˚
sin234˚ + cos234˚ + 2tan18˚ cot 18˚ − cot230˚
(sin234˚ + cos234˚) + 2tan18˚(1/tan18˚) − cot230˚
1 + 2 − (√3)2
3 − 3
0
LHS = RHS
Question 11. Without using trigonometrically tables, evaluate:
cosec2 57° – tan2 33° + cos44° cosec46° – √2 cos45° – tan2 60°
Solution:
cosec2 57° – tan2 33° + cos44° cosec46° – √2 cos45° – tan2 60°
cosec2 (90˚-33°) – tan2 33° + cos44° cosec(90˚-44°) – √2 cos45° – tan2 60°
sec33° – tan2 33° + cos44° sec44° – √2 cos45° – tan2 60°
1 + 1 – √2 cos45° – tan2 60°
1 + 1 – √2 (1/√2) – (√3)2
2 – 1 – 3
-2
LHS = RHS
Exercise 21 D
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:
(i) 21˚
= sin 21˚
= 0.3584
(ii) 34˚ 42’
= sin(34˚ 40’ + 2′)
= 0.5693
(iii) 47˚ 32’
= sin(47˚ 30’ + 2’)
= 0.7373 + 0.0004
= 0.7377
(iv) 62˚ 57’
= sin(62˚ 50’ + 7′)
= 0.8906
(v) 10˚ 20˚ + 20˚ 45’
= sin(30˚ 65’)
= sin(31˚ 5’)
= 0.5150 + 0.0012
= 0.5162
Question 2. Use table to find cosine of:
(i) 2˚ 4’
(ii) 8˚ 12’
(iii) 26˚ 32’
(iv) 65˚ 41’
(v) 9˚ 23’ + 15˚ 54’
Solution:
(i) 2˚ 4’
= cos2˚ 4’
= 0.9994-0.0001
= 0.9993
(ii) 8˚ 12’
= cos8˚ 12’
= 0.9898
(iii) 26˚ 32’
= cos(26˚ 30’ + 2^’)
= 0.8949-0.0003
= 0.8946
(iv) 65˚ 41’
= cos65˚ 41’
= cos(65˚ 36’ + 5′)
= 0.4131-0.0013
= 0.4118
(v) 9˚ 23’ + 15˚ 54’
= cos(9˚ 23’ + 15˚ 54’)
= cos(24˚ 77’)
= cos(25˚ 17’)
= cos(25˚ 12’ + 5′)
= 0.9048-0.0006
= 0.9042
Question 3. Use trigonometrical tables to find tangent of:
(i) 37˚
(ii) 42˚ 18’
(iii) 17˚ 27’
Solution:
(i) 37˚
= tan37˚
= 0.7536
(ii) 42˚ 18’
= tan42˚ 18′
= 0.9099
(iii) 17˚ 27’
= tan(17˚ 27’)
= tan(17˚ 24’ + 3′)
= 0.3134 + 0.0010
= 0.3144
Question 4. Use tables to find the acute angle Ɵ, if the value of sin Ɵ is:
(i) 0.4848
(ii) 0.3827
(iii) 0.6525
Solution:
(i) From the tables,
sin29˚ = 0.4848
Ɵ = 29˚
(ii) From the tables,
sin29˚ = 0.4848
Ɵ = 29˚
(iii) From the table,
sin40˚ 42′ = 0.6521
sinƟ-sin40˚ 42′ = 0.6525-0.6521
sinƟ-sin40˚ 42′ = 0.0004
Ɵ = 40˚ 44′
Question 5. Use tables to find the acute angle Ɵ, if the value of cosƟ is:
(i) 0.9848
(ii) 0.9574
(iii) 0.6885
Solution:
(i) From the table,
cos10˚ = 0.9848
Ɵ = 10˚
(ii) From the table,
cos16˚ 48′ = 0.9574-0.9573 = 0.0001
From the table,
1^’ = 0.0001
Ɵ = 16˚ 48′-1′
Ɵ = 16˚ 47′
(iii) From the table,
cos46˚ 30′ = 0.6885-0.6884 = 0.0001
From the table,
1^’ = 0.0001
Ɵ = 46˚ 30′-1′
Ɵ = 46˚ 29′
Question 6. Use tables to find the acute angle Ɵ, if the value of tan q is:
(i) 0.2419
(ii) 0.4741
(iii) 0.7391
Solution:
(i) From the table,
tan13˚ 36′ = 0.2419
Ɵ = 13˚ 36′
(ii) From the table,
tan25˚ 18′ = 0.4727-0.4727 = 0.0014
From the table
4’ = 0.0014
Ɵ = 25˚ 18′ + 4′
Ɵ = 25˚ 22′
(iii) From the table,
tan36˚ 24′ = 0.7391-0.7373 = 0.0018
From the table
4’ = 0.0018
Ɵ = 36˚ 24′ + 4′
Ɵ = 36˚ 28′
Exercise 21 E
Question 1. If sinA + cosA = p and secA + cosecA = q, the prove that: q(p2−1) = 2p
Solution:
It is given that, q(p2−1) = 2p
(secA + cosecA)[(sinA + cosA)2−1]
We know that (a + b)2 = a2 + b2 + 2ab
= (secA + cosecA)[(sinA + cosA)2−1]
= (secA + cosecA)[sin2A + cos2A + 2sinAcosA − 1]
= (secA + cosecA)[1 + 2sinAcosA−1]
= (secA + cosecA)[2sinAcosA]
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-536.png)
Question 2. If x = acosƟ and y = bcotƟ, show that:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-537.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-538.png)
Question 3. If secA + tanA = p, show that:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-539.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-540.png)
Question 4. If tan A = ntanB and sinA = msinB, prove that:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-541.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-542.png)
Question 5. (i) If 2sinA – 1 = 0, show that: sin3A = 3sinA – 4sin3 A
(ii) If 4cos2 A – 3 = 0, show that: cos3A = 4cos3 A – 3cosA
Solution:
(i) It is given that, 2sinA – 1 = 0
2sinA = 1
sinA = 1/2
We know that, sin30˚ = 1/2
sinA = sin30˚
A = 30˚
From LHS
sin3A
sin3(30˚)
sin90˚
We know that sin 90˚ = 1
sin90˚ = 1
From RHS
= 3sinA – 4sin3 A
= 3sin(30˚) – 4sin3 A
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-543.png)
cosA = cos30˚
A = 30˚
From LHS
cos3A
cos3(30˚)
cos90˚
0
From RHS
= 4cos3 A – 3cosA
= 4cos3 (30˚) – 3cos(30˚)
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-544.png)
Question 6.
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-545.png)
Solution:
![](https://www.icseboards.com/wp-content/uploads/2022/06/image-546.png)
Question 7. If 4cos2 A – 3 = 0 and 0˚ ≤ A ≤ 90˚, then prove that:
(i) sin3A = 3sinA – 4sin3 A
(ii) cos3A = 4cos3 A – 3cosA
Solution:
It is given that, 4cos2 A – 3 = 0
4cos2 A = 3
cos2 A = 3/4
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-547.png)
cosA = cos30˚
A = 30˚
(i) sin3A = 3sinA – 4sin3 A
From LHS
sin3(30˚)
sin90˚
We know that the value of sin90˚ is 1.
sin90˚ = 1
From RHS,
3sinA – 4sin3 A
3sin(30˚) – 4sin3 (30˚)
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-548.png)
ii) cos3A = 4cos3 A – 3cosA
From LHS
cos3(30˚)
cos90˚
We know that, cos90˚ = 0
cos90˚ = 0
From RHS
= 4cos3 (30˚) – 3cos(30˚)
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-549.png)
Question 8. Find A, if 0˚ ≤ A ≤ 90˚ and:
(i) 2cos2 A – 1 = 0
(ii) sin3A – 1 = 0
(iii) 4sin2A – 3 = 0
(iv) cos2A – cosA = 0
(v) 2cos2A + cosA – 1 = 0
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-550.png)
(ii) sin3A – 1 = 0
sin3A = 1
We know that sin90˚ = 1
sin3A = sin90˚
3A = 90˚
A = 90˚/3
A = 30˚
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-551.png)
(iv) cos2 A – cosA = 0
cosA(cosA – 1) = 0
cosA = 0 or cosA – 1 = 0
cosA = 0 or cosA = 1
We know that, cos90˚ = 0 and cos0˚ = 1
cosA = cos90˚ and cosA = cos0˚
A = 90˚ and A = 0˚
So, the value of A is 90˚ or 0˚
(v) 2cos2 A + cosA – 1 = 0
2cos2 A + 2cosA – cosA – 1 = 0
2cosA(cosA + 1) – 1(cosA + 1) = 0
(2cosA – 1)(cosA + 1) = 0
2cosA – 1 = 0 or cosA + 1 = 0
2cosA = 1 or cosA = – 1
cosA = 1/2 or cosA = – 1
We know that, cos60˚ = 1/2
cosA = cos60˚
A = 60˚
Question 9. If 0˚<A<90˚; find A, if:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-552.png)
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-553.png)
Question 10. Prove that: (cosecA – sinA)(secA – cosA)sec2 A = tanA
Solution:
From LHS
(cosecA – sinA)(secA – cosA)sec2 A
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-554.png)
Question 11. Prove the identity (sinƟ + cosƟ)(tanƟ + cotƟ) = secƟ + cosecƟ.
Solution:
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-555.png)
Question 12. Evaluate without using trigonometric tables,
sin2 28° + sin2 62° + tan2 38°– cot2 52° + 1/4 sec2 30°
Solution:
sin2 28° + sin2 62° + tan2 38°–cot2 52° + 1/4 sec2 30°
sin2 28° + [sin(90 – 28)°]2 + tan2 38° – [cot2 (90 – 38)°]2 + 1/4 sec2 30°
sin2 28° + cos2 28˚ + tan2 38° – tan2 38 + 1/4 sec2 30°
![Selina ICSE Class 10 Maths Solutions Chapter 21 Trigonometrical Identities](https://www.icseboards.com/wp-content/uploads/2022/06/image-556.png)