How do you factor and simplify ${tan}^{2} x - {cot}^{2} x$ $tan^2x-cot^2x$ ?

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How do you factor and simplify ${tan}^{2} x - {cot}^{2} x$ $tan^2x-cot^2x$ ?

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Answer 1 · 2016-11-29T13:30:07

$- \frac{4 (sin x - cos x) (sin x + cos x)}{{sin}^{2} 2 x}$ $-(4(sin x - cos x)(sin x + cos x))/(sin^2 2x)$ or
-4cot 2x.csc 2x

Explanation:

${tan}^{2} x - {cot}^{2} x = \frac{{sin}^{2} x}{{cos}^{2} x} - \frac{{cos}^{2} x}{{sin}^{2} x} =$ $tan^2 x - cot^2 x = sin^2 x/(cos^2 x) - cos^2 x/(sin^2 x) =$
$= \frac{{sin}^{4} x - {cos}^{4} x}{{sin}^{2} x . {cos}^{2} x} =$ $= (sin^4 x - cos^4 x)/(sin^2 x.cos ^2 x)=$
Since:
$(a) = ({sin}^{4} x - {cos}^{4} x) = ({sin}^{2} x - {cos}^{2} x) ({sin}^{2} x + {cos}^{2} x) =$ $(a) = (sin^4 x - cos^4 x) = (sin ^2 x - cos^2 x)(sin^2 x + cos^2 x) =$
$= (sin x - cos x) (sin x + cos x)$ $= (sin x - cos x)(sin x + cos x)$ , and
$(b) = {sin}^{2} x . {cos}^{2} x = (\frac{1}{4}) {sin}^{2} 2 x$ $(b) = sin^2 x.cos^2 x = (1/4)sin^2 2x$ ,
There for:
${tan}^{2} x - {cot}^{2} x = \frac{(a)}{(b)} = \frac{4 (sin x - cos x) (sin x + cos x)}{{sin}^{2} 2 x}$ $tan^2 x - cot^2 x = ((a))/((b)) = (4(sin x - cos x)(sin x + cos x))/(sin^2 2x)$

There is another answer for simplification:
Since (sin^2 x - cos^2 x) = - cos 2x, then,
$\frac{(a)}{(b)} = - \frac{4 cos 2 x}{{sin}^{2} 2 x} = - (4 cot 2 x) (\frac{1}{sin 2 x}) =$ $((a))/((b)) = - (4cos 2x)/(sin^2 2x) = - (4 cot 2x)(1/(sin 2x)) =$
$- 4 cot 2 x . csc 2 x$ $- 4cot 2x.csc 2x$ #

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How do you factor and simplify ${tan}^{2} x - {cot}^{2} x$ $tan^2x-cot^2x$ ?

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How do you factor and simplify ${tan}^{2} x - {cot}^{2} x$ $tan^2x-cot^2x$ ?