Даю 89 баллов!Помогите упражненеим 169!

Question 1

Question 2

169.
$a) \frac{1}{ \frac{sin \alpha }{cos \alpha }+ \frac{cos \alpha }{sin \alpha } } = \\ \\ \frac{1}{ \frac{sin ^{2} \alpha+cos ^{2} \alpha}{cos \alpha*sin \alpha}}= \\ \\ \frac{1}{ \frac{1}{cos \alpha*sin \alpha}}=sin \alpha *cos \alpha =0.5*sin2 \alpha$

б) (1-sin²α/cos²α)*cos²α=cos²α-sin²α/cos²α*cos²α=cos²α-sin²α=cos2α

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

$0.5*(2*sin \frac{ \pi - \alpha }{2} *cos \frac{ \pi - \alpha }{2} )= \\ 0.5*sin \frac{2( \pi - \alpha )}{2}=0.5sin( \pi - \alpha )= 0.5sin \alpha$

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

$\frac{4*ctg \frac{ \alpha }{2} }{1-tg^{2} \frac{ \alpha }{2} } = \frac{4*ctg \frac{ \alpha }{2} }{1- \frac{sin ^{2} \frac{ \alpha }{2} }{cos ^{2} \frac{ \alpha }{2} } } = \\ \\ \frac{4*ctg \frac{ \alpha }{2} }{ \frac{cos ^{2} \frac{ \alpha }{2} -sin ^{2} \frac{ \alpha }{2} }{cos ^{2} \frac{ \alpha }{2} } } = \\ \\ \frac{4ctg \frac{ \alpha }{2} *cos ^{2} \frac{ \alpha }{2}}{cos \alpha } = \\$ $\frac{4 \frac{cos \frac{ \alpha }{2} }{sin \frac{ \alpha}{2}} *cos ^{2} \frac{ \alpha }{2}}{cos \alpha } =\\ \frac{2sin \alpha }{cos \alpha } =2tg \alpha$