The principal amplitude of sin40° + icos40°5
70°
- 110°
110°
- 70°
Find the general solution of secθ + 1 = 2 + 3tanθ
Given, secθ + 1 = 2 + 3tanθOn squaring both sides, we get secθ + 12 = 2 + 32tan2θ⇒ secθ + 12 = 4 + 3 + 43sec2θ - 1⇒ secθ + 1secθ + 1 - 7 + 43secθ - 1 = 0⇒ secθ + 18 + 43 - 6 + 43secθ = 0⇒ secθ + 1 = 0or 8 + 43 - 6 + 43 = 0⇒ cosθ = 32 = cosπ6⇒ θ = 2nπ ± π6But θ = 2nπ - π6 does not satisfying the given equation.
∴ θ = 2nπ ± π, 2nπ + π6, n ∈ I
The real part of 1 - cosθ + 2isinθ- 1
13 + 5cosθ
15 - 3cosθ
13 - 5cosθ
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