Question

sin^2x tgx - cos^2 ctgx +2ctg2x=?
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Answer 1

Ответ:

$\sin {}^{2} ( \gamma ) tg (\gamma ) - \cos {}^{2} ( \gamma ) \times ctg( \gamma ) + 2ctg(2 \gamma ) = \\ = \sin {}^{2} ( \gamma ) \times \frac{ \sin( \gamma ) }{ \cos( \gamma ) } - \cos {}^{2} ( \gamma ) \times \frac{ \cos( \gamma ) }{ \sin( \gamma ) } + 2ctg(2 \gamma ) = \\ = \frac{ \sin {}^{3} ( \gamma ) }{ \cos( \gamma ) } - \frac{ \cos {}^{3} ( \gamma ) }{ \sin( \gamma ) } + 2ctg(2 \gamma ) = \\ = \frac{ \sin {}^{4} ( \gamma ) - \cos {}^{4} ( \gamma ) }{ \sin( \gamma ) \cos( \gamma ) } + 2ctg (2\gamma ) = \\ = \frac{( \sin {}^{2} ( \gamma ) - \cos {}^{2} ( \gamma ) )( \sin {}^{2} ( \gamma ) + \cos {}^{2} ( \gamma )) }{ \sin( \gamma ) \cos( \gamma ) } + 2ctg (2\gamma ) = \\ = \frac{ - \cos( 2\gamma ) \times 1 }{ \sin( \gamma ) \cos( \gamma ) } + 2ctg (2\gamma ) = - \frac{2 \cos(2 \gamma ) }{ 2\sin( \gamma ) \cos( \gamma ) } + 2ctg(2 \gamma ) = \\ = - \frac{2 \cos( 2\gamma ) }{ \sin( 2\gamma ) } + 2ctg(2 \gamma ) = - 2ctg(2 \gamma ) + 2ctg(2 \gamma ) = 0$