Question #2f091

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Question #2f091

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Answer 1 · 2016-08-23T18:01:41

We will use the following:

$tan (θ) = \frac{sin (θ)}{cos (θ)}$ $tan(theta) = sin(theta)/cos(theta)$
$sec (θ) = \frac{1}{cos (θ)}$ $sec(theta) = 1/cos(theta)$
${tan}^{2} (θ) + 1 = {sec}^{2} (θ)$ $tan^2(theta)+1 = sec^2(theta)$
$sin (2 θ) = 2 sin (θ) cos (θ)$ $sin(2theta) = 2sin(theta)cos(theta)$

With those:

$\frac{12 tan (a)}{1 + {tan}^{2} (a)} = 6 \frac{2 tan (a)}{1 + {tan}^{2} (a)}$ $(12tan(a))/(1+tan^2(a)) = 6(2tan(a))/(1+tan^2(a))$

$= 6 \frac{2 tan (a)}{{sec}^{2} (a)}$ $=6(2tan(a))/sec^2(a)$

$= 6 \frac{2 tan (a) \cdot {cos}^{2} (a)}{{sec}^{2} (a) \cdot {cos}^{2} (a)}$ $=6(2tan(a)*cos^2(a))/(sec^2(a)*cos^2(a)$

$= 6 \frac{2 \frac{sin (a)}{cos (a)} \cdot {cos}^{2} (a)}{\frac{1}{{cos}^{2} (a)} \cdot {cos}^{2} (a)}$ $=6(2sin(a)/cos(a)*cos^2(a))/(1/cos^2(a)*cos^2(a))$

$= 6 (2 sin (a) cos (a))$ $=6(2sin(a)cos(a))$

$= 6 sin (2 a)$ $=6sin(2a)$

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Question #2f091

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