How do you prove that ${cot}^{2} x + {csc}^{2} x = 2 {csc}^{2} x - {sin}^{2} x - {cos}^{2} x$ $cot^2 x +csc^2 x = 2csc^2 x - sin^2x-cos^2x$ ?

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How do you prove that ${cot}^{2} x + {csc}^{2} x = 2 {csc}^{2} x - {sin}^{2} x - {cos}^{2} x$ $cot^2 x +csc^2 x = 2csc^2 x - sin^2x-cos^2x$ ?

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Answer 1 · 2015-12-20T09:45:03

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

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

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

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

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

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

$= {cot}^{2} (x)$ $= cot^2(x)$

Then, as

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

Adding ${csc}^{2} (x)$ $csc^2(x)$ to both sides gives us

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

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How do you prove that ${cot}^{2} x + {csc}^{2} x = 2 {csc}^{2} x - {sin}^{2} x - {cos}^{2} x$ $cot^2 x +csc^2 x = 2csc^2 x - sin^2x-cos^2x$ ?

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How do you prove that ${cot}^{2} x + {csc}^{2} x = 2 {csc}^{2} x - {sin}^{2} x - {cos}^{2} x$ $cot^2 x +csc^2 x = 2csc^2 x - sin^2x-cos^2x$ ?