How to prove that : $1 - \frac{{cos}^{4} A}{{sin}^{4} A} = 1 + 2 {cot}^{2} A$ $1-cos^4A/sin^4A= 1 + 2cot^2A$ ?

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How to prove that : $1 - \frac{{cos}^{4} A}{{sin}^{4} A} = 1 + 2 {cot}^{2} A$ $1-cos^4A/sin^4A= 1 + 2cot^2A$ ?

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Answer 1 · 2018-03-30T16:20:53

This identity is FALSE

Explanation:

Putting on a common denominator, we get:

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

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

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

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

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

#(2sin^2A - 1)/sin^4A = (sin^2A + 2(1- sin^2A))/sin^2A

$\frac{2 {sin}^{2} A - 1}{{sin}^{4} A} = \frac{2 - {sin}^{2} A}{{sin}^{2} A}$ $(2sin^2A - 1)/sin^4A = (2 - sin^2A)/sin^2A$

$\frac{2}{{sin}^{2} A} - \frac{1}{{sin}^{4} A} = \frac{2}{{sin}^{2} A} - 1$ $2/sin^2A - 1/sin^4A = 2/sin^2A - 1$

And since $\frac{1}{{sin}^{4} A} \neq 1$ $1/sin^4A != 1$ , this identity is false.

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How to prove that : $1 - \frac{{cos}^{4} A}{{sin}^{4} A} = 1 + 2 {cot}^{2} A$ $1-cos^4A/sin^4A= 1 + 2cot^2A$ ?

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Explanation:

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How to prove that : $1 - \frac{{cos}^{4} A}{{sin}^{4} A} = 1 + 2 {cot}^{2} A$ $1-cos^4A/sin^4A= 1 + 2cot^2A$ ?