Question #0166b

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Question #0166b

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Answer 1 · 2017-02-22T07:36:06

$\frac{1 - {sin}^{4} x + {cos}^{4} x}{{cos}^{2} x {sin}^{2} x}$ $( 1-sin^4x+cos^4x) / (cos^2xsin^2x)$

$= \frac{1 - ({sin}^{4} x - {cos}^{4} x)}{{cos}^{2} x {sin}^{2} x}$ $=( 1-(sin^4x-cos^4x)) / (cos^2xsin^2x)$

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

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

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

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

$= \frac{2}{{sin}^{2} x} = 2 {csc}^{2} x$ $=2 / sin^2x=2csc^2x$

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Question #0166b

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