How do you prove ${csc}^{4} [θ] - {cot}^{4} [θ] = 2 {csc}^{2} - 1$ $csc^4[theta]-cot^4[theta]=2csc^2-1$ ?

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Answer 1 · 2016-05-02T21:05:40

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Left Side: $= {csc}^{4} θ - {cot}^{4} θ$ $=csc^4 theta - cot^4 theta$

$= \frac{1}{{sin}^{4} θ} - \frac{{cos}^{4} θ}{{sin}^{4} θ}$ $=1/sin^4 theta - cos^4 theta /sin^4 theta$

$= \frac{1 - {cos}^{4} θ}{{sin}^{4} θ}$ $=(1-cos^4 theta)/sin^4 theta$

$= \frac{(1 + {cos}^{2} θ) (1 - {cos}^{2} θ)}{{sin}^{4} θ}$ $=((1+cos^2 theta)(1-cos^2 theta))/sin^4 theta$

$= \frac{(1 + {cos}^{2} θ) {sin}^{2} θ}{{sin}^{4} θ}$ $=((1+cos^2 theta)sin^2 theta)/sin^4 theta$

$= \frac{1 + {cos}^{2} θ}{{sin}^{2} θ}$ $=(1+cos^2 theta)/sin^2 theta$

$= \frac{1}{{sin}^{2} θ} + \frac{{cos}^{2} θ}{{sin}^{2} θ}$ $=1/sin^2 theta + cos^2 theta/sin^2 theta$

$= {csc}^{2} θ + {cot}^{2} θ$ $=csc^2 theta +cot^2 theta$ ---> ${cot}^{2} θ = {csc}^{2} θ - 1$ $cot^2 theta = csc^2 theta -1$

$= {csc}^{2} θ + {csc}^{2} θ - 1$ $=csc^2 theta+csc^2 theta -1$

$= 2 {csc}^{2} θ - 1$ $=2csc^2 theta -1$

$=$ $=$ Right Side

How do you prove csc^4[theta]-cot^4[theta]=2csc^2-1csc4[θ]−cot4[θ]=2csc2−1csc^4[theta]-cot^4[theta]=2csc^2-1?