How do you prove $\frac{{cos}^{4} x - {sin}^{4} x}{{sin}^{2} x} = {cot}^{2} x - 1$ $(cos^4x - sin^4x)/sin^2x=cot^2x-1$ ?

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How do you prove $\frac{{cos}^{4} x - {sin}^{4} x}{{sin}^{2} x} = {cot}^{2} x - 1$ $(cos^4x - sin^4x)/sin^2x=cot^2x-1$ ?

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Answer 1 · 2015-12-05T22:23:07

Prove trig equation.

Explanation:

${cos}^{4} x - {sin}^{2} x = ({cos}^{2} x - {sin}^{2} x (({cos}^{2} x + {sin}^{2} x) =$ $cos^4 x - sin^2 x = (cos ^2 x - sin^2 x( (cos^2 x + sin^2 x) =$
$= (cos \circ x - {sin}^{2} x) .$ $= (cos@ x - sin^2 x).$ because (cos^2 x + sin^2 x) = 1
$\frac{{cos}^{2} x - {sin}^{2} x}{{sin}^{2} x} = \frac{{cos}^{2} x}{{sin}^{2} x} - \frac{{sin}^{2} x}{{sin}^{2} x} =$ $(cos^2 x - sin^2 x)/(sin^2 x) = cos^2 x/(sin^2 x) - sin^2 x/(sin^2 x) =$
$= {cot}^{2} x - 1$ $= cot^2 x - 1$

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How do you prove $\frac{{cos}^{4} x - {sin}^{4} x}{{sin}^{2} x} = {cot}^{2} x - 1$ $(cos^4x - sin^4x)/sin^2x=cot^2x-1$ ?

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Explanation:

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How do you prove (cos^4x - sin^4x)/sin^2x=cot^2x-1cos4x−sin4xsin2x=cot2x−1(cos^4x - sin^4x)/sin^2x=cot^2x-1?

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Explanation:

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How do you prove $\frac{{cos}^{4} x - {sin}^{4} x}{{sin}^{2} x} = {cot}^{2} x - 1$ $(cos^4x - sin^4x)/sin^2x=cot^2x-1$ ?