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
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
A.
Given, cosθ - α = a, cosθ - β = bWe have,α - β = θ - β - θ - α ...(i)∴ cosα - β = cosθ - βcosθ - α + sinθ - βsinθ - αand sinα - β = sinθ - βcosθ - α - sinθ - αcosθ - β⇒ cosα - β = b . a + 1 - a2 1 - b2and sinα - β = a 1 - b2 - b1 - a2Now, sin2α - β = a 1 - b22 + b1 - a22 - 2ab1 - a2 1 - b2⇒ sin2α - β = a21 - b2 + b21 - a2 - 2abcosα - β - ab⇒ sin2α - β + 2abcosα - β = a2 - a2b2 + b2 - b2a2 + 2a2b2⇒ sin2α - β + 2abcosα - β = a2 + b2