Question

докажите тождество:
$\tan( \gamma ) + \cot( \gamma ) + \tan(3 \gamma ) + \cot(3 \gamma ) = \frac{8 \cos(2 \gamma ) ^{2} }{ \sin(6 \gamma ) }$
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Answer 1

$tg\alpha+Ctg\alpha +tg3\alpha+Ctg3\alpha=(tg\alpha+tg3\alpha) +(Ctg\alpha+Ctg3\alpha)=\\\\=\frac{Sin(\alpha+3\alpha)}{Cos\alpha Cos3\alpha} +\frac{Sin(\alpha+3\alpha)}{Sin\alpha Sin3\alpha} =\frac{Sin4\alpha}{Cos\alpha Cos3\alpha} +\frac{Sin4\alpha}{Sin\alpha Sin3\alpha}=\\\\=\frac{Sin4\alpha(Sin\alpha Sin3\alpha+Cos\alpha Cos3\alpha)}{Sin\alpha Cos\alpha Sin3\alpha Cos3\alpha} =\frac{Sin4\alpha*Cos(3\alpha-\alpha)}{\frac{1}{4}*2Sin\alpha Cos\alpha*2Sin3\alpha Cos3\alpha}=$

$=\frac{4Sin4\alpha Cos2\alpha}{Sin2\alpha Sin6\alpha} =\frac{4*2Sin2\alpha Cos2\alpha Cos2\alpha}{Sin2\alpha Sin6\alpha}=\boxed{\frac{8Cos^{2}2\alpha}{Sin6\alpha}}$