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)
$frac{tg alpha }{tg alpha + ctg alpha } = {sin}^{2} alpha \ frac{sin alpha }{cos alpha } div (frac{sin alpha }{cos alpha } + frac{cosalpha }{sin alpha } ) = {sin}^{2} alpha \ frac{ sinalpha }{cos alpha } divfrac{ {sin}^{2} alpha + {cos}^{2} alpha }{sin alpha times cos alpha } = {sin}^{2} alpha \ frac{ sinalpha }{cos alpha } times frac{ sin alpha times cos alpha }{1 } = {sin}^{2} alpha \ {sin}^{2} alpha = {sin}^{2} alpha$

Answer 2

Спасибо,обожаю вас

Answer 3

ок))