# RD Sharma Solutions Chapter 5 Trigonometric Ratios Exercise 5.3 Class 10 Maths

 Chapter Name RD Sharma Chapter 5 Trigonometric Ratios Book Name RD Sharma Mathematics for Class 10 Other Exercises Exercise 5.1Exercise 5.2 Related Study NCERT Solutions for Class 10 Maths

### Exercise 5.3 Solutions

Evaluate the following :

1. sin 20°/cos 70°

Solution

(ii) cos19°/sin71°

Solution

(iii) sin 21°/cos 69°

Solution

(iv) tan 10°/cot 80°

Solution

(v) sec 11°/cosec 79°

Solution

Evaluate the following :

2. (i) [sin 49°/cos 45°]2 + [cos 41°/sin 49°]

Solution

(ii) cos 48° - sin 42°

Solution

cos 48° = cos(90° - 42°) sin 42°
∴ sin 42° - sin 42° = 0

(iii) cot 40°/cos 35° - 1/2[ cos 35°/sin 55°]

Solution

(iv) [sin 27°/cos 63°] - [cos 63°/sin 27°]2

Solution

(v) tan 35°/ cot 55° + cot 63°/cos 63° - 1

Solution

(vi) sec 70°/cosec 20° + sin 59°/cos 31°

Solution

(vii) sec 50° sin 40° + cos 40° cosec 50°

Solution

sec 50° = sec (90° - 40°) = cosec 40°
cos 40° = cos(90° - 50°) = sin 50°
∴ sin Î¸ cosec Î¸ = 1
⇒ cosec 40° sin 40° + sin 50° cosec 50°
⇒ 1+ 1  = 2

3. Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) sin 59° + cos 56°

Solution

sin 59° = sin (90° - 59°) = cos 31°
cos 56° = cos (65° - 34°) = sin 34°
⇒ cos 31° + sin 34°

(ii) tan 65° + cot 49°

Solution

tan 65° = tan(90° - 25°) = cot 5°
cot 49° = cot(90° - 41°) = tan 41°
⇒ cot 25° + tan 41°

(iii) sec 76° + cosec 52°

Solution

sec 76° = sec(90° - 14°) = cosec 14°
cosec 52° = cosec(90° - 88°) = sec 38°
⇒ cosec 14° + sec 38°

(iv) cos 78° + sec 78°

Solution

cos 78° = cos(90° - 12°) = sin 12°
sec 78° = sec(90° - 12°) = cosec 12°
⇒ sin 12° + cosec 12°

(v) cosec 54° + sin 72°

Solution

cosec 54° = cosec(90° - 36°) = sec 36°
sin 72° = sin(90° - 18°) = cos 18°
⇒ sec 36° + cos 18°

(vi) cot 85° + cos 75°

Solution

cot 85° = cot(90° - 5°) = tan 5°
cos 75° = cos(90° - 15°) = sin 15°
= tan 5° + sin 15°

(vii) sin 67° + cos 75°

Solution

sin 67° = sin(90° - 23°) = cos 23°
cos 75° = cos(90° - 15°) = sin 15°
= cos 23° + sin 15°

4. Express cos 75° + cot 75° in terms of angles between 0° and 30°.

Solution

cot 75° = cos(90° - 15°) = sin 15°
cot 75° = cot(90° - 15°) = tan 15°
= sin 15° + tan 15°

5. If sin3A = cos(A - 26°), where 3A is an acute angle, find the value of A = ?

Solution

Cos Î¸ = sin (90° - Î¸)
⇒ cos(A - 26) = sin(90° - (A - 26°) ]
⇒ sin 3A = sin(90° - (A - 26°)]
Equating angles on both sides
3A = 90° - A + 26°
4A = 116°
A = 116°/4 = 29°
∴ A = 29°

6. If A, B, C are interior angles of a triangle ABC, prove that

(i) tan[(c + a)/2 = cot(B/2)

(ii) sin (B+C)/2 = cos A/2

Solution

7. Prove that

(i) tan 20° tan 35° tan 45° tan 55° tan 70° = 1

Solution

tan 20° = tan(90° - 70°) = cot 70°
tan 35° = tan(90° - 70°) = cot 55°
tan 45° = 1
⇒ cot 70° tan 70° × cot 55° tan 55° × tan 45° cot Î¸ = tan Î¸ = 1
⇒ 1 × 1 × 1 = 1

Hence proved.

(ii) sin 48° sec 42° + cosec 42° = 2

Solution

sin 48° = sin(90° - 42°) = cos 42°
cos 45° = cos(90° - 42°) = sin 42°
sec Î¸ . cos Î¸ = 1 . sin Î¸ cosec Î¸ = 1
⇒ cos 42° sec 42° + sin 42° cosec 42°
⇒ 1 + 1  = 2
∴ LHS = RHS

(iii) sin 70°/cos 20° + cosec 20°/sec 70° - 2 cos 70° cosec 20° = 0

Solution

sin(70°) = sin(90° - 20°) = cos 20°
cosec 20° = cosec(90° - 70°) = sec 70°
cos 70° = cos(90° - 20°) = sin 20°
⇒ cos 20°/cos 20° + sec 70°/sec 70° - 2 sin20° cosec 20°
1 + 1 - 2(1) = 0
∴ LHS = RHS

Hence proved

(iv) cos 80°/sin 10° + cos 59° cosec 31° = 2

Solution

cos 80° = cos(90° - 10°) = sin 10°
cos 59° = cos(90° - 31°) = sin 31°
⇒ sin 10°/sin 10° + sin 31° cosec 31°
= 1 + 1 = 2  [∵ sin Î¸ cosec Î¸ = 1]
Hence proved

8. Prove the following .

(i) sin Î¸ sin(90 - Î¸) - cosÎ¸ cos(90 -Î¸) = 0

Solution

sin(90 - Î¸) = cos Î¸
cos(90 - Î¸) - cos Î¸ sin Î¸  = 0
∴ LHS = RHS
Hence proved

(ii) [cos(90° - Î¸) sec(90° - Î¸) tanÎ¸]/[cosec(90° - Î¸) sin(90° - Î¸) cot(90° - Î¸) + tan(90° - Î¸)/cotÎ¸ = 2

Solution

cos(90° - Î¸) = sinA
cosec(90° - Î¸) = sec Î¸
sec(90° - Î¸) = cosec Î¸
sin(90° - Î¸) = cos Î¸
cot(90° - Î¸) = tan Î¸

(i
ii) [tan(90° - A)cotA]/cosec2 A - cos2 A = 0

Solution
tan(90 - A) = cot A

(iv) [cos(90° - A) sin(90° - A)]/tan(90° - A) - sin2 A = 0
Solution

(v) sin(50° + Î¸) - cos(40° - Î¸) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1
Solution
sin(50° + Î¸ ) = cos(90° - (50° + Î¸) = cos(40° - Î¸)
tan 1 = tan(90° - 89°) . cot 89°
tan 10° = tan(90° - 80°) = cot 80°
tan 20° = tan(90° - 70°) = cot 70°
⇒ cos(40° - Î¸) - cos(40° - Î¸) = cots 89° tan 89°. cot 80° cot 70° tan 70°
cot. tanÎ¸ = 1
= 1 . 1 . 1 = 1
LHS = RHS
Hence proved

9. Evaluate :
(i) 2/3 (cos4 30° - sin4 45°) - 3(sin2 60° - sec2 45°) + 1/4 cot2 30°
Solution

(ii) 4(sin2 30° + cos4 60°) - 2/ 3 [{√(3/2)}2 ・{1/√2}2 } + 1/4 (√3)2
Solution

(iii) sin 50°/cos 40° + cosec 40°/sec 50° - 4cos 50° cosec 40°
Solution

(iv) tan 35° tan 40° tan 50° tan 55°
Solution
tan 35° = tan (90° - 55°) = cot 55°
tan 40° = tan(90° - 50°) = cot 55°
tan 65° = 1
cot 55° tan 55° cot 50° tan 50° tan 45°
1 × 1× 1 = 1

(v) cosec(65° + Î¸) - sec(25° -Î¸) - tan(55° - Î¸) + cot(35° + Î¸)
Solution
cosec(65° + Î¸) = sec(90° - (65° + Î¸)) = sec(25° - Î¸)
tan(55° - Î¸) = cot(90° - (55° - Î¸) = cot(35° + Î¸)
⇒ sec(25° - Î¸) - sec(25° -Î¸) tan (55° - Î¸) + tan(55° - Î¸) = 0

(vi) tan 7° tan23° tan60° tan 67° tan83°
Solution
tan 7° tan 23° tan 60° tan(90° - 23) tan(90° - 7°)
⇒ tan 7° tan 23° tan 60° cot 23° tan 60°
1×1× √3 = √3

(vii) 2sin 68°/cos 22° - 2 cot 15°/5 tan 75° - (8 tan 45° tan 20° tan40° tan 50° tan 70°)/5
Solution
Sin 68° = sin(90° - 22°) = cos 22°
Cot 15° = tan(90° - 75°) = tan 75°

= 2 - 2/5 - 3/5 = 2 - 1 = 1

(viii) 3 cos 55°/ 7 sin 35° - 4(cos 70° cosec 20°)/7(tan 5° tan 25° tan 45° tan 65° tan 85°)
Solution

(ix) sin 18°/cos 72° + √3[tan 10° tan 30° tan 40° tan 50° tan 80°]
Solution

(x) cos 58°/sin 32° + sin 22°/cos 68° - (cos 38° cosec 52°)/(tan 18° tan 35° tan 60° tan 72° tan 65°)
Solution

10. If  sin Î¸ = cos(Î¸ - 45°), where Î¸ - 45° are acute angles, find the degree measure of Î¸.
Solution
Sin Î¸ = cos (Î¸ - 45°)
Cos Î¸ = cos (90 - Î¸)
Cos (Î¸ - 45°) = sin(90° - (Î¸ - 45°)) = sin(90° - Î¸ + 45°)
Sin Î¸ = sin (135 - Î¸)
Î¸ = 135 - Î¸
2Î¸ = 135
∴ Î¸ = 135/2

11. If A, B, C are the interior angles of a Î”ABC, show that :
(i) Sin (B+C)/2 = cos A/2
(ii) cos (B+C)/2 = Sin A/2
Solution
A + B +C = 180
B - C = 180 - A/2

12. If 2Î¸ + 45° and 30° - Î¸ are acute angles, find the degree measure of Î¸ satisfying Sin(20 + 45°) = cos(30 - Î¸).
Solution
Here 20 + 45° and  30 - Î¸° are acute angles:
We know that (90 - Î¸) = cos Î¸
Sin (2Î¸ + 45°) = sin (90 - (30 - Î¸))
Sin(2Î¸ + 45°) = sin (90° - 30 + Î¸)
Sin(20 + 45°) = sin( 60 + Î¸)
On equating sin of angle of we get
2Î¸ + 45 = 60 + Î¸
2Î¸ - Î¸ = 60 - 45
Î¸ = 15°

13. If Î¸ is a positive acute angle such that sec Î¸ = cosec 60° , find 2 cos2Î¸ – 1.
Solution
We know that sec(90 - Î¸) = cosec2Î¸
Sec Î¸ = sec (90 - 60°)
On equating we get
sec Î¸ = sec 30°
Î¸ = 30°

14. If cos 2Î¸ = sin 4Î¸ where 2Î¸, 4Î¸ are acute angles, find the value of Î¸.
Solution
We know that sin(90 - Î¸) = cos Î¸
sin 2Î¸ = cos 2Î¸
⇒ sin 4Î¸ = sin(90 - 2Î¸)
⇒ 4Î¸ = 90 - 2Î¸
⇒ 6Î¸ = 90
⇒ Î¸ = 90/6
⇒ Î¸ = 15°

15. If Sin 3Î¸ = cos(Î¸ - 6°) where 3 Î¸ and Î¸ - 6° are acute angles, find the value of Î¸.
Solution
3Î¸, Î¸ - 6° are acute angle
We know that sin(90 - Î¸) = cos Î¸
sin 3Î¸ = sin(90 - (Î¸ - 6°))
⇒ sin 3Î¸ = sin(90 - Î¸ + 6°)
⇒ sin 3Î¸ = sin (96° - Î¸)
⇒ 3Î¸ = 96° - Î¸
⇒ 4Î¸ = 96°
⇒ Î¸ = 96°/4
⇒ Î¸ = 24°

16. If sec 4A = cosec(A - 20°) where 4A is acute angle, find the value of A.
Solution
sec 4A = sec [90 − A − 20] [∵ sec(90 − Î¸) = cosec Î¸]
⇒ sec 4A = sec (90 – A + 20)
⇒ sec 4A = sec (110 – A)
⇒ 4A = 110 – A
⇒ 5A = 110
⇒ A = 110/5 = 22

17. If  sec 2A = cosec (A - 42°) where 2A  is acute angle. Find the value of A.
Solution
We know that [sec(90 – Î¸)) = cosec Î¸
sec 2A = sec (90 – (A – 42))
⇒ sec 2A = sec (90 – A + 42)
⇒ sec 2A = sec (132 – A)
Now equating both the angles we get
2A = 132 – A
⇒ 3A = 132/3
⇒ A = 44