The number of solution of tan5πcosθ = cot

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 Multiple Choice QuestionsMultiple Choice Questions

781.

If f : R  R be the signum function and g : R  R be the greatest integer function, then sinπfog12 is equal to

  • 1

  • 32

  • 0

  • 12


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782.

The number of solution of tan5πcosθ = cot5πsinθ for θ in 0, 2π will be

  • 28

  • 14

  • 4

  • 2


D.

2

We have,tan5πcosθ = cot5πsinθ for θ in 0, 2π tan5πcosθ = tanπ2 - 5πsinθ 5πcosθ + 5πsinθ = π2 cosθ + sinθ = 110On multiplying both sides by 12, we getcosθ2 + sinθ2 = 1102 cosθsinπ4 + sinθcosπ4 = 1102 sinπ4 + θ = 1102       sinA + B = sinAcosB + cosAsinB only 2 solutions are there in (0, 2π).


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783.

AB is a vertical pole. The end A is on the ground level. C is the middle point of AB and P is a point on the ground level. The portion BC subtends an angle β at P. If AP = nAB, then tanβ is equal to

  • n2n2 + 1

  • nn2 + 1

  • nn + 1

  • None of these


784.

tanθ + cotθ = 2 = 2, then sinθ is equal to

  • 12

  • 13

  • 12

  • 1


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785.

If θ = π6, then the 10th term of  1 + cosθ + isinθ + cosθ + isinθ2 + cosθ + isinθ3 + ...is equal to

  • i

  • - 1

  • 1

  • - i


786.

sin5θsinθ is equal to 

  • 16cos4θ - 12cos2θ + 1

  • 16cos4θ+ 12cos2θ + 1

  • 16cos4θ - 12cos2θ - 1

  • 16cos4θ + 12cos2θ - 1


787.

cos2π6 + θ - sin2π6 + θ is equal to

  • 12cos2θ

  • 0

  • - 12cos2θ

  • 12


788.

In ABC, cosC + cosAc + a + cosBb is equal to

  • 1a

  • 1b

  • c + ab

  • 1


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789.

In ABC, ab2 - c2 + cb2 - a2 = 0, then B is equal to 

  • π2

  • π4

  • 2π3

  • π3


790.

In ABC,a2sin(2C) + c2sin(2A) is equal to 

  • 2

  • 3

  • 4


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