Помогите решить 2sin^2А-1/1+2cos^2A - cos^2A+1/1-cos^2A. Created by 5454117. matematika-ru

1-cos^2A...

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2sin^2А-1/1+2cos^2A - cos^2A+1/1-cos^2A

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mshatun

$\frac{2 { \sin }^{2} \alpha - 1}{1 + 2 { \cos}^{2} \alpha } - \frac{ { \cos}^{2} \alpha + 1 }{1 - { \cos }^{2} \alpha } = \frac{2 { \sin }^{2} \alpha - { \cos }^{2} \alpha - { \sin}^{2} \alpha }{(1 + 2 { \cos}^{2} \alpha)(1 - { \cos }^{2} \alpha)} - \frac{ { \cos }^{2} \alpha + { \cos }^{2} \alpha + { \sin}^{2} \alpha }{(1 + 2 { \cos}^{2} \alpha)(1 - { \cos }^{2} \alpha)} = \frac{ { \sin }^{2} \alpha - { \cos}^{2} \alpha }{(1 + 2 { \cos}^{2} \alpha)(1 - { \cos }^{2} \alpha)} - \frac{2 { \cos }^{2} \alpha + { \sin}^{2} \alpha }{(1 + 2 { \cos}^{2} \alpha)(1 - { \cos }^{2} \alpha)} = \frac{ - 3 { \cos}^{2} \alpha }{1 + 2 { \cos}^{2} \alpha - { \cos}^{2} \alpha - 2 { \cos}^{4} \alpha } = \frac{ - 3 { \cos}^{2} \alpha }{ { \sin}^{2} \alpha + { \cos }^{2} \alpha + { \cos}^{2} \alpha - 2 { \cos }^{4} \alpha } = \frac{ - 3 { \cos}^{2} \alpha}{{ \sin}^{2} \alpha + 2 { \cos }^{2} \alpha - 2 { \cos }^{4} \alpha}$

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