Question

Find tan2θ if θ terminates in Quadrant IV and cosθ = 3/5.

A. -1/2
B. 1/2
C. 24/7
D. -24/7

Answer 1

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`\cos( \alpha ) = (3)/(5) \\`

`{sin}^(2)( \alpha ) = 1 - {cos}^(2) ( \alpha )`

`{sin}^(2)( \alpha ) = 1 - ({ (3)/(5) })^(2) \\`

`{sin}^(2)( \alpha ) = 1 - (9)/(25) \\`

`{sin}^(2)( \alpha ) = (25)/(25) - (9)/(25) \\`

`{sin}^(2)( \alpha ) = (16)/(25) \\`

`sin( \alpha )= ± \sqrt{ (16)/(25) } \\`

In Quadrant IV , sin is negative .

Thus ;

`sin( \alpha ) = - \sqrt{ (16)/(25) } \\`

`\sin( \alpha ) = - (4)/(5) \\`

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`\tan(2 \alpha ) = ( \sin(2 \alpha ) )/( \cos(2 \alpha ) ) \\`

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`\sin(2 \alpha ) = 2. \sin( \alpha ). \cos( \alpha )`

`\sin(2 \alpha ) = 2 * ( - (4)/(5) ) * ( (3)/(5) ) \\`

`\sin(2 \alpha ) = - (24)/(25) \\`

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`\cos(2 \alpha ) = {cos}^(2)( \alpha ) - {sin}^(2)( \alpha )`

`\cos(2 \alpha ) = ({ (3)/(5) })^(2) - ({ - (4)/(5) })^(2) \\`

`\cos(2 \alpha ) = (9)/(25) - (16)/(25) \\`

`\cos(2 \alpha ) = - (7)/(25) \\`

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`\tan(2 \alpha ) = ( \sin(2 \alpha ) )/( \cos(2 \alpha ) ) \\`

`\tan(2 \alpha ) = ( - (24)/(25) )/( - (7)/(25) ) \\`

`\tan(2 \alpha ) = - (24)/(25) / - (7)/(25) \\`

`\tan(2 \alpha ) = - (24)/(25) * - (25)/(7) \\`

`\tan(2 \alpha ) = (24)/(7) \\`

Thus the correct answer is (( C )) .

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Answer 2

we have

cos θ = 3/5.

sin²θ=1-cos²θ

sin²θ=1-9/25

sin²θ=16/25

sinθ=±4/5

since it lies in IV qyadrant so

sinθ=-4/5

we have

tan2θ=sin2θ/cos2θ=2sinθcosθ/[cos²θ-sin²θ)

=(2×-4/5×3/5)/(9/25-16/25)=24/7