Question

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Answer 1

Ответ:

$\frac{1 - {ctg}^{2} ( \alpha - \frac{3\pi}{2} )}{ctg( \alpha + \frac{\pi}{2}) } \times \frac{tg( \alpha - \frac{\pi}{2} )}{1 - {ctg}^{2} ( \alpha - 2\pi)} = \\ = \frac{1 - {tg}^{2} ( \alpha )}{ - tg( \alpha )} \times \frac{( - ctg( \alpha ))}{1 - {ctg}^{2} ( \alpha )} = \\ = \frac{1 - \frac{ { \sin }^{2} ( \alpha )}{ { \cos }^{2} (\alpha )} }{ - tg( \alpha )} \times \frac{( - ctg( \alpha ))}{1 - \frac{ { \cos }^{2} ( \alpha )}{ { \sin}^{2}( \alpha ) } } = \\ = \frac{ctg( \alpha )}{tg( \alpha )} \times \frac{ { \cos }^{2} ( \alpha ) - { \sin }^{2} (\alpha )}{ { \cos }^{2} ( \alpha )} \times \frac{ { \sin }^{2} (\alpha )}{ { \sin}^{2} (\alpha ) - { \cos}^{2} (\alpha ) } = \\ = \frac{ctg( \alpha )}{tg( \alpha )} \times {tg}^{2} ( \alpha ) \times \frac{( - ( { \sin }^{2} (\alpha ) - { \cos }^{2} (\alpha) )) }{ { \sin}^{2} (\alpha ) - { \cos}^{2}( \alpha ) } = \\ = ctg( \alpha ) \times tg( \alpha ) \times ( - 1) = \\ = - \frac{1}{tg( \alpha )} \times tg( \alpha ) = - 1$