Помогите прошу...171-172 задания

Question 1

Question 2

Номер 171
1) $sin^{2} \alpha +cos^{2} \alpha +tg ^{2} \alpha - \frac{1}{cos^{2} \alpha } =sin^{2} \alpha +cos^{2} \alpha + \frac{sin ^{2} \alpha }{cos^{2} \alpha } - \frac{1}{cos^{2} \alpha}$ $=1+ \frac{sin^{2} \alpha-1}{cos^{2} \alpha} = \frac{cos^{2}\alpha+sin^{2}\alpha-1}{cos^{2}\alpha} =sin^{2}\alpha-1=-cos^{2} \alpha$
2) $2- \frac{1}{sin^{2} \alpha } +ctg^{2} \alpha =2- \frac{1}{sin^{2} \alpha } + \frac{cos^{2} \alpha }{sin^{2} \alpha } = \frac{2sin^{2} \alpha -1+cos ^{2} \alpha }{sin^{2} \alpha } = 2-1$ $+cos^{2} \alpha =cos^{2} \alpha +1$
Номер 172
1) $sin^{2} \alpha=1-cos^{2} \alpha =1- \frac{1}{4}= \frac{3}{4}$
$sin\alpha= \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3}}{2}$
$tg=\frac{sin \alpha }{cos \alpha } = \frac{ \frac{ \sqrt{3}}{2} }{ \frac{1}{2} } = \frac{ 2\sqrt{3} }{2}= \sqrt{3}$
2) $cos x^{2} =1-sin^{2} =1- \frac{3}{4}= \frac{1} }{4}$
$cos \alpha = \sqrt{ \frac{1}{4} } = \frac{1}{2}$
$tg \alpha = \frac{sin \alpha }{cos \alpha } = \frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } = \frac{2 \sqrt{3} }{2} = \sqrt{3}$
3) $cos^{2} \alpha =1-sin^{2} \alpha =1- \frac{1}{16} = \frac{15}{16}$
$cos \alpha = \sqrt{ \frac{15}{16} } = \frac{ \sqrt{15} }{4}$
$tg \alpha= \frac{sin \alpha }{cos \alpha }= \frac{ \frac{1}{4} }{ \frac{ \sqrt{15} }{4} }= \frac{4}{4 \sqrt{15} } = \frac{1}{ \sqrt{15} } = \frac{ \sqrt{15} }{15}$