Question #5a859

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Question #5a859

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Answer 1 · 2016-08-14T13:31:47

We will start from the left hand side and show that it equals the right hand side. To do so, we will use the following identities:

$tan (x) = \frac{sin (x)}{cos (x)}$ $tan(x) = sin(x)/cos(x)$
$sec (x) = \frac{1}{cos (x)}$ $sec(x) = 1/cos(x)$
${sec}^{2} (x) - 1 = {tan}^{2} (x)$ $sec^2(x) - 1 = tan^2(x)$

Note that the third identity may be derived from ${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x) + cos^2(x) = 1$ by dividing each side by ${cos}^{2} (x)$ $cos^2(x)$ and then subtracting $1$ $1$ from each side.

Proceeding:

$L H S = {tan}^{2} (u) - {sin}^{2} (u)$ $LHS = tan^2(u) - sin^2(u)$

$= {(\frac{sin (u)}{cos (u)})}^{2} - {sin}^{2} (u)$ $= (sin(u)/cos(u))^2-sin^2(u)$

$= \frac{{sin}^{2} (u)}{{cos}^{2} (u)} - {sin}^{2} (u)$ $= sin^2(u)/cos^2(u)-sin^2(u)$

$= {sin}^{2} (u) (\frac{1}{{cos}^{2} (u)} - 1)$ $=sin^2(u)(1/cos^2(u) - 1)$

$= {sin}^{2} (u) ({sec}^{2} (u) - 1)$ $=sin^2(u)(sec^2(u)-1)$

$= {sin}^{2} (u) {tan}^{2} (u) = R H S$ $=sin^2(u)tan^2(u) = RHS$ ∎

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Question #5a859

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