Question

если sinA=-40/41,tgB=9/40,A∈Ⅳч,то чему равен tg(a+b)

Answer 1

sinα = -40/41; tgβ = 9/40

cosα = √(1 - cos²α) = √(1 - 40²/41²) = √((41² - 40²)/41²) = √((81)/41²) = 9/41;

tgα = sinα/cosα = -40/9

tg(α+β) = (tgα + tgβ)/(1 - tgα · tgβ) = (-40/9 + 9/40)/(1 + 40/9 · 9/40) = -1519/(2 · 360) = -1519/720

Answer 2

sina=-40/41
tgb=9/40 ;a€|V
tg(a+b)=(tga+tgb)/(1-tga*tgb)=?

cos²a=1-sin²a=1-1600/1681=81/1681
cosa=9/41;a€|V

tga=sina/cosa=-40/41:9/41=-40/9
tg(a+b)=(-40/9+9/40)(1+9/40*40/9)
=(-1600+81)/(2*360)=
-1519/720