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

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

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Answer 1 · 2016-04-22T12:20:59

Proof below

Explanation:

Identity: ${sec}^{2} θ = 1 + {tan}^{2} θ$ $sec^2theta=1+tan^2theta$
${sec}^{2} (\frac{π}{2} - x) - 1 = 1 + {tan}^{2} (\frac{π}{2} - x) - 1$ $sec^2(pi/2-x)-1=1+tan^2(pi/2-x)-1$
$= {tan}^{2} (\frac{π}{2} - x)$ $=tan^2(pi/2-x)$
Identity: $tan (\frac{π}{2} - θ) = cot θ$ $tan(pi/2-theta)=cottheta$
$= {cot}^{2} x$ $=cot^2x$

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

1 Answer

Explanation:

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How do you verify sec^2((pi/2)-x)-1=cot^2 xsec2((π2)−x)−1=cot2xsec^2((pi/2)-x)-1=cot^2 x?

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Explanation:

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