Question

Помогите с №В4. ПЛИЗ

Answer 1

1+tg²α=1/cos²α
sin²α+cos²α=1, cos²α=1-sin²α
$cos ^{2} alpha =1- ( frac{ sqrt{17} }{17} )^{2} cos^{2} alpha = frac{16}{17}$
$1+ tg^{2} alpha =1: frac{16}{17} tg^{2} alpha = frac{17}{16} -1 tg^{2} alpha = frac{1}{16}$

tgα=√(1/16). tgα=1/4. tgα=0,25

ответ: tgA=0,25