Question # 1

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1 + cos Θ

1 - cos Θ

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quad I

quad. II

quad. III

quad. IV

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quad. I

quad. II

quad. III

quad. IV

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csc (2π - Θ), where Θ is a basic angle, will have terminal side in:

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quad. I

quad. II

quad. III

quad. IV

sin (Θ - π) =

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tan (270° + Θ) is equal:

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cot Θ

tan Θ

-cot Θ

-tan Θ

Cot1°, Cot2°, Cot3°, .......... Cot89° =

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-1

1

∞

none

2 cos α sin ß =

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cos (α + ß) + cos (α - ß)

sin (α + ß) + sin (α - ß)

sin (α + ß) - sin (α - ß)

cos (α + ß) + cos (α - ß)

tan (α+ß) =

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The angles 90°±Θ, 180°±Θ, 270°±Θ, 360°±Θ, are the:

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composite angles

half angles

quadrantal angles

allied angles

tan (-135°) =

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0

1

√2

sin5Θ + sin3Θ is equal to:

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2cos 2Θ sin Θ

-2 cos 4Θ sin Θ

-2 sin 4Θ cos Θ

2 sin 4Θ cos Θ

If sin α = cos ß in any triangle ABC then:

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α + ß = 90°

α + ß = 180°

α + ß = 360°

α + ß

-2 sin α sin ß =

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sin (α + ß) + sin (α - ß)

cos (α + ß) + cos (α - ß)

cos (α + ß) - cos (α - ß)

cos (α - ß) + cos (α - ß)

cos (α-ß) =

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cos α cos ß + sin α sin ß

cos α cos ß - sin α sin ß

cos α cos ß + sin α cos ß

sin α cos ß - sin α sin ß

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2 cos α cos ß =

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sin (α + ß) - sin (α - ß)

cos (α + ß) - cos (α - ß)

cos (α + ß) + cos (α - ß)

sin (α + ß) + sin (α - ß)

tan (α - ß ) =

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2 sin 12° sin 46° =

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cos 34° + cos 58°

sin 34° - sin 58°

sin 34° + sin 58°

cos 34° - cos 58°

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