R.D. Sharma Solutions Class 10th: Ch 5 Trigonometric Ratios MCQ

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 θ is an acute angle such that cos θ = 3/5, then sin θ tan θ - 1/2 tan2θ = 
(i) 16/625
(ii) 1/36
(iii) 3/160
(iv) 160/3

Solution


2. If tan θ = a/b, then a sin θ + b cos θ/a sin θ - b cos θ 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 θ - 4 = 0, then the value of 5 sin θ - 4 cos θ/5 sin θ + 4 cos θ 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 θ = 1/√7 , then cosec2 θ - sec2 θ/cosec2 θ - sec2 θ =   
(a) 5/7
(b) 3/7
(c) 1/12
(d) 3/4

Solution


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

Solution


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

Solution


9. If 3 cos θ = 5 sin θ, then the value of 5 sin θ - 2 sec3 θ+2 cos θ/5 sin θ + 2 sec3 θ-2 cos θ 
(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° - θ) cot (90° - θ) , 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 θ is an acute angle such that sec2 θ = 3, then the value of tan2 θ-cosec2 θ/tan2 θ + cosec2 θ 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 cotθ tanθ = 1]
= 1

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

Solution

We have to find : cos(90°-θ)sec(90°-θ)tanθ/cosec(90°-θ)sin(90°-θ)cot(90°-θ) + tan(90°-θ)/cot θ
so cos(90°-θ)sec(90°-θ)tanθ/cosec(90°-θ) sin(90°-θ) cot(90°-θ) + tan(90°-θ)/cot θ
= sinθ cosecθ tanθ/secθ cosθ tanθ + cotθ/cotθ
= 1×tanθ/1×tanθ + cotθ/cotθ
= 1+1
= 2

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

Solution

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

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

Solution

We are given that 5θ and 4θ are acute angles satisfying the following condition
sin 5θ = cos 4θ . we are asked to find 2 sin 3θ - √3tan 3θ
⇒ sin 5θ = cos 4θ
⇒ cos(90°-5θ) = cos 4θ
⇒ 90° = 5θ = 4θ
⇒ 9θ = 90°
Where 5θ and 4θ are acute angles
⇒ θ = 10°
Now we have to find :
2 sin 3θ - √3 tan 3θ
= 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 θ = 2/3, then 2 sec2 θ + 2 tan2 θ - 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°-θ)= cotθ
tanθ cotθ = 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 φ  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 ,
θ + φ + 90° = 180°
so, θ = 90° - ...(1)
Now from the triangle ?ABC,
sin θ  = 4/5
Now we will use equation (1) in the above,
sin(90° - φ ) = 4/5
Therefore, cos φ = 4/5

35. In fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot  θ
(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 θ = CB/AB
= 12/5 

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