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
The number of solution of tan5πcosθ = cot5πsinθ for θ in 0, 2π will be
28
14
4
2
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
A.
We have, AC = BC = 12AB,∠BPC = β, AP = nABLet, ∠APC = αIn ∆APC,tanα = ACAP = 12ABnAB = 12nIn ∆APB,tanα + β = ABAP = ABnAB = 1n∴ tanα + β - γ = tanα + β - tanα1 + tanα + βtanα ∵ tanx - y = tanx - tany1 + tanxtany⇒ tanβ = 1n - 12n1 + 1n12n⇒ tanβ = n2n2 + 1
tanθ + cotθ = 2 = 2, then sinθ is equal to
13
If θ = π6, then the 10th term of 1 + cosθ + isinθ + cosθ + isinθ2 + cosθ + isinθ3 + ...is equal to
i
- 1
- i
sin5θsinθ is equal to
16cos4θ - 12cos2θ + 1
16cos4θ+ 12cos2θ + 1
16cos4θ - 12cos2θ - 1
16cos4θ + 12cos2θ - 1
cos2π6 + θ - sin2π6 + θ is equal to
12cos2θ
- 12cos2θ
In ∆ABC, cosC + cosAc + a + cosBb is equal to
1a
1b
c + ab
In ∆ABC, ab2 - c2 + cb2 - a2 = 0, then B is equal to
π2
π4
2π3
π3
In ∆ABC,a2sin(2C) + c2sin(2A) is equal to
∆
2∆
3∆
4∆