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Q1. Expression representing sin 15° is

• 4sin³ 5° - 3 sin 5°
• 3sin 5° - 4 sin³ 5°
• 4cos³ 5° - 3cos 5°
• 3cos 5° - 4cos³ 5°

Solution

We know that sin 3x = 3 sin X - 4 sin³ X Substituting X = 5° We get sin 15° = 3 sin 5° - 4 sin³ 5°

Q2. Sin5x-Sinx expressed as a product is

•   2sin3xcos2x
•   2cos3xsin2x
•   2sin2xcos3x
•   2cos2xsin3x

Solution

  We know that sinA-sinB= 2cos((A+B)/2)sin((A-B)/2) Using this and substituting, A=5x and B= x we get Sin5x-Sinx=2 cos3xsin2x

Q3. In the III quadrant value of sin x

• increases from 0 to 1
• increases from -1 to 0
• increases from -∞ to 0
• decreases from 0 to -1

Solution

In the III quadrant value of sin x decreases from 0 to -1.

Q4. Which of the following statement is incorrect

• sec90^o is defined.
• sec270^o is not defined.
• tan90^o is not defined.
• cot0^o is not defined.

Solution

Since cos90^o = 0, so sec90^o is not defined.

Q5. If  cos x = cosy So, x =

• 2nπ± y
• π + y
• π - y
• π± y

Solution

cos x - cos y = 0,Applying the formula we get, We get ,hence x=2nπ + y or x = 2nπ - y Hence, the answer.

Q6. The value of cosx in the second quadrant

• increases from  0 to 1
• decreases from 1 to 0
• increases from -1 to 0
• decreases from 0 to -1

Solution

The value of cos function at 90 degrees is 0 and at 180 degrees it’s -1.Thus as the angle moves from 90 degrees to 180 degrees i.e. in the second  quadrant , the value decreases from 0 to -1.

Q7. What is the sign of the sinA and tanA in third quadrant respectively

•     positive and negative 
•     negative and positive
•     positive and positive          
•     negative and negative.

Solution

  Follows from the ASTC rule in III quadrant  only tan is positive

Q8. The most general valuie that satisfies the equation cosecθ = 2 and cotθ = -√3 is

• 2nπ + π/6, n ∈Z
• 2nπ - 5π/6, n ∈ Z
• 2nπ + 5π/ 6, n ∈ Z
• 2nπ ± 5π/ 6, n ∈ Z

Solution

sinθ = ½  and tanθ = -1/√3 ⇒ θ lies in the second quadrant sinθ = ½ = sin π/6 = sin (π-π/6) = sin5π/6 and tanθ = -1/ √3= tan (π-π/6) = tan 5π/6 ⇒ θ = 2nπ + 5π/6, n ∈ Z is the general solution

Q9. What is the range of cos function?

•   [0,1]
• [-1,0]
•   [-1,1]
•   [-2,2]

Solution

  Range of sin and cos functions is [-1,1]

Q10. In which quadrant are sin, cos and tan positive?

•   Ist quadrant
•   IInd  quadrant 
•   IIIrd  quadrant 
•     IVth quadrant 

Solution

  Sin, cos and tan are all positive in first quadrant.

Q11. A cyclist travels at the rate of 14.4 km/hr and the radius of the wheel is 35 cm. What is the measure of the angle in radian through which a spoke of the wheel will turn in one second?

•     41/5 radians
•     70/8 radians
•     80/7 radians
•     40/3 radians

Solution

  The distance traveled in 1 hour= 14.4 km/hr = 1440000 cm/hr =1440000/3600 cm /sec =400 cm in 1 sec the formula for the length of the arc, when the angle in radians and radius are given is using this formula and the given values =400/35=80/7 radians   

Q12. Find the value of cos18⁰+sin36⁰

• 2cos36⁰sin18⁰
• - 2cos36⁰sin18⁰
• 2sin36⁰cos18⁰
• -  2sin36⁰cos18⁰

Solution

sin36⁰=cos(90⁰-36⁰) = cos54⁰ So,cos18⁰+sin36⁰=cos18⁰+cos54⁰=2cos 36⁰sin(-18⁰)= -2cos36⁰sin18⁰

Q13. sin (n+1)x cos(n+2)x-cos(n+1)x sin(n+2)x=

• sinx
• -sinx
• cosx
• -cos x

Solution

sinA cosB-cosA sinB=sin(A-B) So , substituting, A=(n+1)x and B=(n+2)x  ,we get the answer So , substituting, A=(n+1)x and B=(n+2)x  ,we get sin [(n+1)x – (n+2)x] = sin (-x) = -sinx

Q14. If cos A¸ is negative , so A¸ lies in quadrants

• I  and II
• II  and III
• III and IV
• IV and I

Solution

Cosine ratio corresponds to X coordinate and X - axis has negative values in quadrants II and III. So cosine ratio in negative in II and III quadrants.

Q15. The period of the function y = tan x is

• π
• 2π
• π
• not defined

Solution

Function tan x is periodic with period π

Q16. A cow is tethered to a corner of a field with a rope of length 7 m.If she grazes on the length of 210 m , what is the angle through which the rope moves?

•     45 radians
•     30 radians
•     60 radians
•     90 radians

Solution

    The formula for the length of the arc, when the angle and radius are given is     so using this formula and substituting the values, we get, 210=7  Hence, =30 radians

Q17. tan x = -5/12, x lies in the second quadrant. So sinx=?

• -5/13
• 5/13
• -12/13
• 12/13

Solution

x lies in the second quadrant so sinx must be positive. Using the identities 1+=  sin x=1/cosec x  As x lies in the second quadrant so sinx must be positive. So we get sinx = 5/13

Q18. Law of cosine can be applied to

• only right angled triangle
• equilateral triangle
• Isosceles triangle
• All types of triangle

Solution

Law of cosines can be applied to all types of triangles.

Q19. The quadrant in which sin θ is negative and tan θ is positive is

• I
• II
• III
• IV

Solution

In III quadrant, sin θ is negative and tan θ is positive.

Q20. Which of the following is correct:

• cos(A + B) = cosA cosB + sinA sinB
• cos(A + B) = cosA + cosB
• cos(A – B) = cosA – cosB
• cos(A – B) = cosA cosB + sinA sinB

Solution

cos(x + y) = cosx cosy – sinx siny Replacing y by –y in the above identity, we get cos (x – y) = cosx cosy + sinx siny

Q21. An object is moving on the circle of radius 7 cm in clockwise direction with a speed of 10 cm/s. find the angle in radians covered by it in 10 sec when it starts at 0 degree.

• 14.29 radian
• 1.43 radian
• -1.43 radian
• -14.29 radian

Solution

 In 1 sec an object travels 10 cm Therefore in 10 sec it travels 100 cm. Therefore, length of arc is 100 cm Since the object is travelling in clockwise direction therefore it would be negative Thus, Angle in radian = -l/r = -100/7 = -14.29 radian  

Q22. cos9y-cos5y=

• 2 cos 7ycos2y
• 2sin 7y sin 2y
• -2cos7y cos 2y
• -2 sin 7y sin 2y

Solution

cosA-cos B= -2sin[(A+B)/2] sin [(A-B)/2] So substituting, A =9y and B=5y, we get cos9y-cos5y = -2sin7y sin2y

Q23. What is the radius of the circle in which a central angle of 60^o intercepts an arc length of 48.4 cm?

• 6.6 cm.
• 46.2 cm
• 64.2 cm
• 2.1 cm

Solution

1^o = radians  So 60^o =  radians We know that l = r                         =46.2 cm

Q24. In the third quadrant the value of the sine function

•   increases from 0 to 1
•   increases from -1 to 0
•   decreases from  1 to 0
•   decreases from 0 to -1

Solution

   The value of sine function at180 degrees is 0 and at 270 degrees it"s -1.Thus as the angle moves from 180 degrees to 270 degrees i.e. in the third quadrant , the value decreases from 0 to -1.

Q25. In a triangle ABC, sin 2A=?

• 2sin(B+C)cos( B-C)
• 2sin( B-C)cos(B+C)
• 2sin(B-C)cos( B-C)
•   -2sin(B+C)cos( B+C)

Solution

In a triangle ABC, we have, A+B+C= So,sin2A=2sinAcosA =2sin[180-(B+C)] cos[ 180-(B+C)]=2sin(B+C) x [-cos(B+C)]=-2sin(B+C) cos(B+C)

Q26. The values of sin105° is equal to

• sin60°
• sin45°
• cos15°
• cos45°

Solution

sin105° = sin(90° + 15°) = cos15°

Q27. The general solution of sin = 0 is

• π
• 2π
• nπ, where n is an integer
• nπ where n is a real number

Solution

If sin x = 0 Then x = nπ for any integer n

Q28. Which of the following is correct:

• sin(x + y) = sinx siny + cos x cosy
• sin(x + y) = sinx + siny
• sin(x – y) = sinx – siny
• sin(x + y) = sinx cosy + cosx siny

Solution

cos (x – y) = cosx cosy + sinx Siny sin(x + y) = =   cos(- x) cosy + sin(- x) siny =   sin x cosy + cos x siny

Q29. What is the sign of the sec θ and cosec θ in second quadrant respectively?

• positive and negative         
• negative and positive
• positive and positive          
• negative and negative

Solution

Follows from the ASTC rule in II quadrant only sin and cosec are positive rest all are negative.  

Q30. Find the central angle of the circle whose diameter is 20 cm subtended by an arc of length 15.4 cm.

• 0.77 radian
• .154 radian
• 1.54 radian
• 15.4 radian

Solution

Q31. A pendulum 36 cm long oscillates through an angle of 10 degrees. Find the length of the path described by its extremity.

•   6.280 cm
•   3.620 cm
•   62.50 cm
•   31.25cm

Solution

  The formula for the length of the arc, when the angle in radians and radius are given is     so using this formula and substituting the values of r=36, we get, l =36 x =2 x (3.14) = 6.28 cm

Q32. A train is travelling on a curve of 350 m radius at 7 Km/hr. Through what angle will it turn in one minute?

• 50 radian
• 1/20 radian
• 3 radian
• 1/3 radian

Solution

In 60 min train travels   7000 m Therefore, in 1 min train travels 7000/60 m So angle in radians = radians

Q33. In a triangle ABC, cosA- cosB=

•   -2cos(C/2) sin(A-B)/2
•   2cos(C/2) sin(A+B/2)
•   - 2sin(C/2) sin(A-B/2)
•   -2sin(C/2) sin(A+B/2)

Solution

  We know that,  cosX -cosY=-2sin((X+Y)/2) sin((X-Y)/2) so  substituting, X=A and  Y=B and the fact that, A+B=180-C, so sin(A+B/2)=sin(180-C/2)=sin(90-{C/2})=cos(C/2) We get, cosA- cosB=-2cos(C/2) sin(A-B/2)  

Q34. What is the value of sin4x=?

• sin2xcos2x
• –sin2xcos2x
• 2 sin2xcos2x
• -2sin2xcos2x

Solution

We know that sin 2A= 2sinA cosA sin4x = 2sin2xcos2x

Q35. Which of the following is positive?

• sin 240^o
• cos 330^o
• tan 315^o
• sec 120^o

Solution

We know that 330^o is in IV quadrant. In IV quadrant, only cosx and secx are positive.

Q36. If sin B¸ is positive, then B¸ lies in the quadrants

•  I  and II 
• II  and III
• III and IV
• IV and I

Solution

Sine ratio corresponds to y coordinate and y - axis is positive in I and II quadrants. So, sine ratio is positive in I and II quadrants.

Q37. 2(bc cos A+ ca cos B + ab cos C)=

• a²+b²+c²
• 2abc
• a²+b²-c²
• abc

Solution

Q38. sin(60° + A) cos(30° - B) + cos(60° + A) sin(30° - B) is equal to

• sin(A – B)
• cos(A – B)
• sin(A + B)
• cos(A + B)

Solution

By using, sin(x + y) = sinx cosy + cosx siny sin[ (60° + A) + (30° - B) ] = sin[90° + ( A - B)]  = cos(A – B)

Q39. What is the length of arc of a circle whose radius is 14 cm and which subtends an angle of 135^o at the centre?

• 42 cm
• 33 cm
• 22 cm
• 28 cm

Solution

Q40. Identify the odd one out from the following

•   sec²θ + tan² θ = 1 
•   sin² θ + cos² θ=1
•   cosec² θ = 1 + cot² θ            
•   sec² θ = 1 + tan² θ

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