Question

помогите решить 1 и 3!!!! ДАЮ 50 баллов

Answer 1

1)
$sin \alpha = \frac{8}{17} \\ cos \alpha = \sqrt{1 - {sin}^{2} \alpha } \\ cos \alpha = \sqrt{1 - { (\frac{8}{17}) }^{2} } = \sqrt{ 1 - \frac{64}{289} } = \\ = \sqrt{ \frac{25}{289} } = \frac{5}{17}$
$tg \alpha = \frac{sin \alpha }{cos \alpha } \\ tg \alpha = \frac{8}{17} \div \frac{5}{17} = \frac{8}{17} \times \frac{17}{5} = \frac{8}{5} \\ \\ ctg \alpha = \frac{1}{tg \alpha } \\ ctg \alpha = \frac{1}{ \frac{8}{5} } = \frac{5}{8}$
2)
а)
$\frac{1 - {sin}^{2} \alpha }{sin \alpha \times cos \alpha } \times tg \alpha = \\ = \frac{ {cos}^{2} \alpha }{sin \alpha \times cos \alpha} \times \frac{sin \alpha }{cos \alpha } = 1$
б)
$\frac{1 - sin \alpha }{cos \alpha } - \frac{cos \alpha }{1 + sin \alpha} = \\ = \frac{(1 - sin \alpha)(1 + sin \alpha) - {cos}^{2} \alpha }{cos \alpha (1 + sin \alpha)} = \\ = \frac{1 - {sin}^{2} \alpha - {cos}^{2} \alpha }{cos \alpha (1 + sin \alpha)} = \\ = \frac{1 - ({sin}^{2} \alpha + {cos}^{2} \alpha )}{cos \alpha (1 + sin \alpha)} = \\ = \frac{1 - 1}{cos \alpha (1 + sin \alpha)} = 0$
в)
$( {cos}^{2} \alpha + \frac{1}{1 + {ctg}^{2} \alpha } ) = {cos}^{2} \alpha + {sin}^{2} \alpha = 1$
3)
$\frac{tg \alpha }{tg \alpha + ctg \alpha } = {sin}^{2} \alpha \\ \frac{sin \alpha }{cos \alpha } \div (\frac{sin \alpha }{cos \alpha } + \frac{cos\alpha }{sin \alpha } ) = {sin}^{2} \alpha \\ \frac{ sin\alpha }{cos \alpha } \div\frac{ {sin}^{2} \alpha + {cos}^{2} \alpha }{sin \alpha \times cos \alpha } = {sin}^{2} \alpha \\ \frac{ sin\alpha }{cos \alpha } \times \frac{ sin \alpha \times cos \alpha }{1 } = {sin}^{2} \alpha \\ {sin}^{2} \alpha = {sin}^{2} \alpha$