25 балов!! Упростить: (〖cos 〗^2 (-α))/(1+sin⁡〖(-α)〗 ) tg(π+α)-ctg( 3π/2-α)Найдите...

$\frac{cos^2( -\alpha )}{1+sin(- \alpha )} = \frac{cos^2( \alpha )}{1-sin( \alpha )} = \frac{1-sin^2( \alpha )}{1-sin( \alpha )} =\frac{[1-sin( \alpha )]*[1+sin( \alpha )]}{1-sin( \alpha )} =\\= 1+sin( \alpha )$
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$tg(\pi+ \alpha )-ctg( \frac{3\pi}{2} - \alpha )=tg( \alpha )-tg( \alpha )=0$
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$6cos^2( \alpha )-3sin^2( \alpha )=6cos^2( \alpha )-3(1-cos^2( \alpha ))=\\= 6cos^2( \alpha )-3+3cos^2( \alpha )=9cos^2( \alpha )-3$

$-1 \leq cos( \alpha ) \leq 1\\ 0 \leq cos^2( \alpha ) \leq 1\\ 0 \leq 9cos^2( \alpha ) \leq 9\\ 0-3 \leq 9cos^2( \alpha )-3 \leq 9-3\\ -3 \leq 9cos^2( \alpha )-3 \leq 6\\\\ -3 \leq 6cos^2( \alpha )-3sin^2( \alpha ) \leq 6$