How do you prove ${sec}^{2} x - {cot}^{2} (\frac{π}{2} - x) = 1$ $sec^2 x - cot^2 ( pi/2-x) =1$ ?

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How do you prove ${sec}^{2} x - {cot}^{2} (\frac{π}{2} - x) = 1$ $sec^2 x - cot^2 ( pi/2-x) =1$ ?

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Answer 1 · 2016-02-05T22:32:46

Using the following:

$sec (x) = \frac{1}{cos (x)}$ $sec(x) = 1/cos(x)$
$cot (x) = \frac{cos (x)}{sin (x)}$ $cot(x) = cos(x)/sin(x)$
$cos (- x) = cos (x)$ $cos(-x) = cos(x)$
$sin (- x) = - sin (x)$ $sin(-x) = -sin(x)$
$cos (x - \frac{π}{2}) = sin (x)$ $cos(x-pi/2) = sin(x)$
$sin (x - \frac{π}{2}) = - cos (x)$ $sin(x-pi/2) = -cos(x)$
${sin}^{2} (x) + {cos}^{2} (x) = 1 \Rightarrow 1 - {sin}^{2} (x) = {cos}^{2} (x)$ $sin^2(x)+cos^2(x) = 1 => 1 - sin^2(x) = cos^2(x)$

We have

${sec}^{2} (x) - {cot}^{2} (\frac{π}{2} - x) = {sec}^{2} (x) - {cot}^{2} (- (x - \frac{π}{2}))$ $sec^2(x)-cot^2(pi/2-x) = sec^2(x) - cot^2(-(x-pi/2))$

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

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

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

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

$= \frac{{cos}^{2} (x)}{{cos}^{2} (x)}$ $=cos^2(x)/cos^2(x)$

$= 1$ $= 1$

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How do you prove ${sec}^{2} x - {cot}^{2} (\frac{π}{2} - x) = 1$ $sec^2 x - cot^2 ( pi/2-x) =1$ ?

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