## Chapter 5 Trigonometric Ratios R.D. Sharma Solutions for Class 10th Math MCQ's

Multiple Choice Questions

Mark the correct alternative in each of the following :

1. If Î¸ is an acute angle such that cos Î¸ = 3/5, then sin Î¸ tan Î¸ - 1/2 tan2Î¸ =
(i) 16/625
(ii) 1/36
(iii) 3/160
(iv) 160/3

Solution

2. If tan Î¸ = a/b, then a sin Î¸ + b cos Î¸/a sin Î¸ - b cos Î¸ is equal to
(a) a2+b2/a2-b2
(b) a2-b2/a2+b2
(c) a+b/a-b
(d) a-b/a+b

Solution

3. If 5 tan Î¸ - 4 = 0, then the value of 5 sin Î¸ - 4 cos Î¸/5 sin Î¸ + 4 cos Î¸ is
(a) 5/3
(b) 5/6
(c) 0
(d) 1/6

Solution

4.If 16 cot x = 12, then sin x - cos x/sin x + cos x equals
(a) 1/7
(b) 3/7
(c) 2/7
(d) 0

Solution

5. If 8 tan x = 15, then sin x - cos x is equal to
(a) 8/17
(b) 17/7
(c) 1/17
(d) 7/17

Solution

6. If tan Î¸ = 1/√7 , then cosec2 Î¸ - sec2 Î¸/cosec2 Î¸ - sec2 Î¸ =
(a) 5/7
(b) 3/7
(c) 1/12
(d) 3/4

Solution

7. If tan Î¸ = 3/4, then cos2 Î¸ - sin2 Î¸ =
(a) 7/25
(b) 1
(c) -7/25
(d) 4/25

Solution

8. If Î¸ is an acute angle such that tan2 Î¸ = 8/7, then the value of (1+sin Î¸)(1-sin Î¸)/(1+cos Î¸)(1-cos Î¸) is
(a) 7/8
(b) 8/7
(c) 7/4
(d) 64/49

Solution

9. If 3 cos Î¸ = 5 sin Î¸, then the value of 5 sin Î¸ - 2 sec3 Î¸+2 cos Î¸/5 sin Î¸ + 2 sec3 Î¸-2 cos Î¸
(a) 271/979
(b) 316/2937
(c) 542/2937
(d) None of these

Solution

10. If tan2 45° - cos2 30° = x sin 45° cos 45°, then x =
(a) 2
(b) -2
(c) -(1/2)
(d) 1/2

Solution

11. The value of cos2 17° - sin2 73° is
(a) 1
(b) 1/3
(c) 0
(d) -1

Solution

12. The value of cos3° 20° - cos3 70°/ sin3 70° - sin3 20° is
(i) 1/2
(ii) 1/√2
(iii) 1
(iv) 2

Solution

13. If x cosec2 30° sec2 45°/ 8 cos2 45° sin2 60°
(a) 1
(b) -1
(c) 2
(d) 0

Solution

14. If A and B are complementary angles , then
(a) sin A = sin B
(b) cos A = cos B
(c) tan A = tan B
(d) sec A = cosec B

Solution

15. If x sin (90° - Î¸) cot (90° - Î¸) , then x =
(a) 0
(b) 1
(c) -1
(d) 2

Solution

16. If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
(a) 1
(b) √3
(c) 1/2
(d) 1/√2

Solution

17. If angles A,B,C to a △ABC from an increasing AP, then sin B =
(a) 1/2
(b) √3/2
(c) 1
(d) 1/√2

Solution

18. If Î¸ is an acute angle such that sec2 Î¸ = 3, then the value of tan2 Î¸-cosec2 Î¸/tan2 Î¸ + cosec2 Î¸ is
(a) 4/7
(b) 3/7
(c) 2/7
(d) 1/7

Solution

19. The value of tan 1° tan 2° tan 3° ... tan 89° is
(a) 1
(b) -1
(c) 0
(d) None of these

Solution

20. The value of cos 1° cos 2° cos 3° ... cos 180° is
(a) 1
(b) -1
(c) 0
(d) None of these

Solution

Here we have to find: cos 1° cos 2° cos 3° ...cos 180°
cos 1° cos 2° cos 3° ...cos 180°
= cos 1° cos 2° cos 3°...cos 89° cos 90° cos 91° ...cos 180° [since cos 90° = 0]
= cos 1° cos 2° cos 3°...0 × cos 90° ...cos 180°
= 0

21.The value of tan 10° tan 15° tan 75° tan 80° is
(a) -1
(b) 0
(c) 1
(d) None of these

Solution

Here we have to find : tan 10° tan 15° tan 75° tan 80°
Now
tan 10° tan 15° tan 75° tan 80°
= tan (90° - 80°) tan (90° - 75°)tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80°
= (cot 80° tan 80°)(cot 75° tan 75°)
= 1×1 [since cotÎ¸ tanÎ¸ = 1]
= 1

22. The value of cos(90°-Î¸) sec(90°-Î¸) tan Î¸/cosec (90°-Î¸) sin (90°-Î¸) cot (90°-Î¸)  + tan (90°-Î¸)/cot Î¸ is
(a) 1
(b) -1
(c) 2
(d) -2

Solution

We have to find : cos(90°-Î¸)sec(90°-Î¸)tanÎ¸/cosec(90°-Î¸)sin(90°-Î¸)cot(90°-Î¸) + tan(90°-Î¸)/cot Î¸
so cos(90°-Î¸)sec(90°-Î¸)tanÎ¸/cosec(90°-Î¸) sin(90°-Î¸) cot(90°-Î¸) + tan(90°-Î¸)/cot Î¸
= sinÎ¸ cosecÎ¸ tanÎ¸/secÎ¸ cosÎ¸ tanÎ¸ + cotÎ¸/cotÎ¸
= 1×tanÎ¸/1×tanÎ¸ + cotÎ¸/cotÎ¸
= 1+1
= 2

23. If Î¸ and 2 Î¸ - 45° are acute angles such that sin Î¸ = cos (2Î¸ - 45°), then tan Î¸ is equal to
(a) 1
(b) -1
(c) √3
(d) 1/√3

Solution

Given that : sin Î¸ = cos (2Î¸ - 45°) and Î¸ and 2Î¸ - 45 are acute angles
we have to find tan Î¸
⇒ sin Î¸ = cos(2Î¸ - 45°)
⇒ cos (90°- Î¸) = cos(2Î¸ - 45°)
⇒ 90°- Î¸ = 2Î¸ - 45°
⇒ 3Î¸ = 135°
Where Î¸ and 2Î¸ - 45° are acute angles
since Î¸ = 45°
Now
tan Î¸
= tan 45° put Î¸ = 45°
= 1

24. If 5Î¸ and 4Î¸ are acute angles satisfying sin 5Î¸ = cos 4Î¸, then 2 sin 3Î¸ - √3 tan 3Î¸ is equal to
(a) 1
(b) 0
(c) -1
(d) 1+√3

Solution

We are given that 5Î¸ and 4Î¸ are acute angles satisfying the following condition
sin 5Î¸ = cos 4Î¸ . we are asked to find 2 sin 3Î¸ - √3tan 3Î¸
⇒ sin 5Î¸ = cos 4Î¸
⇒ cos(90°-5Î¸) = cos 4Î¸
⇒ 90° = 5Î¸ = 4Î¸
⇒ 9Î¸ = 90°
Where 5Î¸ and 4Î¸ are acute angles
⇒ Î¸ = 10°
Now we have to find :
2 sin 3Î¸ - √3 tan 3Î¸
= 2 sin 30° - √3 tan 30°
= 2 × 1/2 - √3 × 1/√3
= 1-1
= 0

25. If A + B = 90°, then tan A tan B + tan A cot B/sin A sec B - sin2 B/cos2 A is equal to
(a) cot2A
(b) cot2B
(c) -tan2A
(d) -cot2A

Solution

26. 2 tan 30°/1+tan230° is equal to
(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°

Solution

27. 1-tan2 45°/1+tan2 45° is equal to
(a) tan 90°
(b) 1
(c) sin 45°
(d) sin 0°

Solution

28.  Sin 2A = 2 sin A is true when A =
(a) 0°
(b) 30°
(c) 45°
(d) 60°

Solution
We are given, sin2A = 2sinA.cosA
so
⇒ sin 2A = 2sinA
⇒ 2 sinA.cosA = 2sinA
⇒ cos A = 1
As A = 0°

29. 2 tan 30°/1-tan2 30° is equal to
(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°

Solution

30. If A,B and C are interior angles of a triangle ABC, then sign (B+C/2) =
(a) sin A/2
(b) cos A/2
(c) - sin A/2
(d) - cos A/2

Solution

We know that in triangle ABC
A+B+C =
⇒ B+C = 180° - A
⇒ B+C/2 = 90°/2 - A/2
⇒ sin (B+C/2) = sin(90°-A/2)
Since sin(90°-A) = cos A
So
sin (B+C/2) = cos A/2

31. If cos Î¸ = 2/3, then 2 sec2 Î¸ + 2 tan2 Î¸ - 7 is equal to
(a) 1
(b) 0
(c) 3
(d) 4

Solution

32. tan 5° × tan 30° × 4 tan 85° is equal to
(a) 4/√3
(b) 4√3
(c) 1
(d) 4

Solution

We have to find tan5° × tan30° × 4 tan85°
We know that
tan(90°-Î¸)= cotÎ¸
tanÎ¸ cotÎ¸ = 1
tan 30° = 1/√3
so
tan5° × tan30° × 4 tan85°
= tan(90°- 85°) × tan30° × 4 tan85°
= cot 85° × tan30° × 4 tan85°
= 4 cot85° × tan85° tan30°
= 4 × 1 × 1/√3
= 4/√3

33. The value of tan 55°/cot35° + cot 1° cot 2° cot 3°... cot 90°, is
(a) -2
(b) 2
(C) 1
(d) 0

Solution

We have to find the value of the following expression
tan 55°/cot 35° + cos 1° cot 2° cot 3° ...cot 90°
= tan 55°/cot 35° + cos 1° cot 2° cot 3° ...cot 90°
= tan(90°-35°)/cot 35° + cot(90°-89°)cot(90°-88°)cot(90°-87°)...cot87 cot88°cot89°...cot90°
= cot35°/cot35° + tan 89° tan 88° tan 87°...cot87 cot 88° cot 89°...cot 90°
= 1+1×1×1...× 0
= 1
As cot 90° = 0

34.  In Fig. 5.47, the value of cos Ï†  is
(a) 5/4
(b) 5/3
(c) 3/5
(d) 4/5

Solution

We should proceed with the fact that sum of angles on one side of a straight line is 180° .
so from the given figure ,
Î¸ + Ï† + 90° = 180°
so, Î¸ = 90° - ...(1)
Now from the triangle ?ABC,
sin Î¸  = 4/5
Now we will use equation (1) in the above,
sin(90° - Ï† ) = 4/5
Therefore, cos Ï† = 4/5

35. In fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot  Î¸
(a) 12/5
(b) 5/12
(c) 13/12
(d) 12/13

Solution

We have the following given data in the figure, AD = 4 cm, BD = 3 cm, CB = 12 cm
Now we will use pythagoras theorem in ABD,
AB = √32 + 42
5 cm
Therefore
cot Î¸ = CB/AB
= 12/5