If sin(A) + sin(B) + sin(C) = 0 and cos(A) + cos(B) + cos(C) = 0, then cos (A + B) + cos (B + C) + cos(C + A) is equal to
cos(A + B + C)
2
1
0
If tanθ . tan120° - θtan120° + θ = 13, then θ is equal to
nπ3 + π18, n ∈ Z
nπ3 + nπ12, n ∈ Z
nπ12 + π12, n ∈ Z
nπ3 + π6, n ∈ Z
1 + cosπ8 - isinπ81 + cosπ8 + isinπ88 = ?
- 1
12
If 1 + tanα1 + tan4α = 2, α ∈ 0, π16,then α = ?
π20
π30
π40
π50
If cosθ = cosα - cosβ1 - cosαcosβ, then one of the values of tanθ2 is
cotβ2tanα2
tanα2tanβ2
tanβ2cotα2
tan2α2tan2β2
The value of the expression 1 + sin2αcos2α - 2πtanα - 3π4 - 14sin2αcotα2 + cot3π2 + α2 is
sin2α2
sin2α
If 16sinθ, cosθ and tanθ are in geometric progression, then the solution set of θ is
2nπ ± π6
2nπ ± π3
nπ + - 1nπ3
nπ + π3
In ∆ABC if x = tanB - C2tanA2, y = tanC - A2tanB2 and z = tanA - B2tanC2, then x + y + z = ?
xyz
- xyz
2xyz
12xyz
B.
Given, x = tanB - C2tanA2 y = tanC - A2tanB2 and z = tanA - B2tanC2⇒ x = b - cb + c ∵tanB - C2 = b - cb + ccotA2Y = c - ac + a and z = a - ba + bNow, 1 + x1 - x = b + c + b - cb + c - b + c By componendo and dividendo⇒ 1 + x1 - x = bcSimilarly,1 + y1 - y = ca and 1 + z1 - z = ab∴ 1 + x1 - x1 + y1 - y1 + z1 - z = bc × ca × ab = 1⇒ 1 + x1 + y1 + z = 1 - x1 - y1 - z⇒ 1 + x + y + z + xy + yz + zx + xyz = 1 - x + y + z + xy + yz + zx - xyz⇒ x + y + z = - xyz
If A > 0, B > 0 and A + B = π3, then the maximum value of AtanB is
13
3
In ∆ABC, if bcosθ = c - a, (where θ is an acute angle), then (c - a) tanθ = ?
2cacosB2
2acsinB2
2casinB2