In ∆ABC if x = tanB - C2

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 Multiple Choice QuestionsMultiple Choice Questions

911.

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


912.

If tanθ . tan120° - θtan120° + θ = 13, then θ is equal to

  • 3 + π18, n  Z

  • 3 + 12, n  Z

  • 12 + π12, n  Z

  • 3 + π6, n  Z


913.

1 + cosπ8 - isinπ81 + cosπ8 + isinπ88 = ?

  • 1

  • - 1

  • 2

  • 12


914.

If 1 + tanα1 + tan4α = 2, α  0, π16,then α = ?

  • π20

  • π30

  • π40

  • π50


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915.

If cosθ = cosα - cosβ1 - cosαcosβ, then one of the values of tanθ2 is

  • cotβ2tanα2

  • tanα2tanβ2

  • tanβ2cotα2

  • tan2α2tan2β2


916.

The value of the expression 1 + sin2αcos2α - 2πtanα - 3π4 - 14sin2αcotα2 + cot3π2 + α2 is

  • 0

  • 1

  • sin2α2

  • sin2α


917.

If 16sinθ, cosθ and tanθ are in geometric progression, then the solution set of θ is

  • 2 ± π6

  • 2 ± π3

  •  +  - 1nπ3

  •  + π3


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918.

In ABC if x = tanB - C2tanA2, y = tanC - A2tanB2 and  z = tanA - B2tanC2, then x + y+ z = ?

  • xyz

  • - xyz

  • 2xyz

  • 12xyz


B.

- xyz

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


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919.

If A > 0, B > 0 and A + B = π3, then the maximum value of AtanB is

  • 13

  • 13

  • 12

  • 3


920.

 In ABC, if bcosθ = c - a, (where θ is an acute angle), then (c - a) tanθ = ?

  • 2cacosB2

  •  2acsinB2

  •  2cacosB2

  • 2casinB2


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