Question #b5658

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Question #b5658

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Answer 1 · 2017-05-25T01:18:34

See Below.

Explanation:

Identities Required:
$sin (2 x) = 2 sin (x) cos (x)$ $sin(2x)=2sin(x)cos(x)$

${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x)=1$
${sin}^{2} (x) = 1 - {cos}^{2} (x)$ $sin^2(x)=1-cos^2(x)$

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

Now solving:
$\frac{sin (2 x)}{sin (x)} - \frac{cos (2 x)}{cos (x)} = sec (x)$ $sin(2x)/sin(x)-cos(2x)/cos(x)=sec(x)$
$\frac{2 sin (x) cos (x)}{sin (x)} - \frac{2 {cos}^{2} (x) - 1}{cos (x)} = R H S$ $(2sin(x)cos(x))/sin(x)-(2cos^2(x)-1)/cos(x)=RHS$
$2 cos (x) - 2 cos (x) + \frac{1}{cos (x)} = R H S$ $2cos(x)-2cos(x)+1/cos(x)=RHS$
$sec (x) = R H S$ $sec(x)=RHS$
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Question #b5658

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