Question

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

Answer 1

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$\cos( \alpha ) = \frac{3}{5}$

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

${sin}^{2}( \alpha ) = 1 - ({ \frac{3}{5} })^{2}$

${sin}^{2}( \alpha ) = 1 - \frac{9}{25}$

${sin}^{2}( \alpha ) = \frac{25}{25} - \frac{9}{25}$

${sin}^{2}( \alpha ) = \frac{16}{25}$

$sin( \alpha )= ± \sqrt{ \frac{16}{25} }$

In Quadrant IV , sin is negative .

Thus ;

$sin( \alpha ) = - \sqrt{ \frac{16}{25} }$

$\sin( \alpha ) = - \frac{4}{5}$

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$\tan(2 \alpha ) = \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) }$

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

$\sin(2 \alpha ) = 2 \times ( - \frac{4}{5} ) \times ( \frac{3}{5} )$

$\sin(2 \alpha ) = - \frac{24}{25}$

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

$\cos(2 \alpha ) = ({ \frac{3}{5} })^{2} - ({ - \frac{4}{5} })^{2}$

$\cos(2 \alpha ) = \frac{9}{25} - \frac{16}{25}$

$\cos(2 \alpha ) = - \frac{7}{25}$

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$\tan(2 \alpha ) = \frac{ \sin(2 \alpha ) }{ \cos(2 \alpha ) }$

$\tan(2 \alpha ) = \frac{ - \frac{24}{25} }{ - \frac{7}{25} }$

$\tan(2 \alpha ) = - \frac{24}{25} \div - \frac{7}{25}$

$\tan(2 \alpha ) = - \frac{24}{25} \times - \frac{25}{7}$

$\tan(2 \alpha ) = \frac{24}{7}$

Thus the correct answer is (( C )) .

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Answer 2

Answer:

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