The principal amplitude of sin40° + icos40°5
70°
- 110°
110°
- 70°
Find the general solution of secθ + 1 = 2 + 3tanθ
The real part of 1 - cosθ + 2isinθ- 1
13 + 5cosθ
15 - 3cosθ
13 - 5cosθ
15 + 3cosθ
D.
Given, 1 - cosθ + 2isinθ- 1
= 2sin2θ2 + i . 4sinθ2cosθ2- 1= 2sinθ2- 1sinθ2 + i2cosθ2- 1= 2sinθ2- 11sinθ2 + i2cosθ2 × sinθ2 - i2cosθ2sinθ2 - i2cosθ2= 2sinθ2- 1sinθ2 - i2cosθ2sin2θ2 + 4cos2θ2
= sinθ2 - i2cosθ22sinθ2sin2θ2 + 4cos2θ2= sinθ2 - i2cosθ22sinθ21 + 3cos2θ2
It's real part= sinθ22sinθ21 + 3cos2θ2= 121 + 3cosθ + 12= 12 + 3cosθ + 3= 15 + 3cosθ
The value of sin36°sin72°sin108°sin144° is equal to
14
116
34
516
If sinA = 110 and sinB = 15 where A and B are positive acute angles, then A + B is equal to
π
π2
π3
π4
The expression tan2α + cot2α is
≥ 2
≤ 2
≥ - 2
None of these
The equation 3sin2x + 10cosx - 6 is satisfied, if
x = nπ ± cos-113
x = 2nπ ± cos-113
x = nπ ± cos-116
x = 2nπ ± cos-116
If ecosx - e- cosx = 4, then the value of cosx is
log2 + 5
- log2 + 5
log- 2 + 5
If tan-1ax + tan-1bx = π2, then x is equal to
ab
2ab
If cosθ - α = a, cosθ - β = b, then
sin2α - β + 2abcosα - β is equal to
a2 + b2
a2 - b2
b2 - a2
- a2 - b2