Question

много баллов!!! помогите, плиз!!!​

Answer 1

$\frac{sin^3a+sin3a}{cos^3a-cos3a}=\frac{sin^3a-4sin^3a+3sina}{cos^3a-4cos^3a+3cosa}=\frac{-3sin^3a+3sina}{-3cos^3a+3cosa}=\frac{3sina(-sin^2a+1)}{3cosa(-cos^2a+1)}=\\ \\ = \frac{3sina*cos^2a}{3cosa*sin^2a}=\frac{cosa}{sina}=ctga$

Answer 2

Воспользуемся формулами тройного угла:

sin3α = 3sinα - 4sin³a

cos3α = 4cos³α - 3cosα

(sin³α + sin3α)/(cos³α - cos3α) = (sin³α + 3sinα - 4sin³α)/(cos³α - 4cos³α + 3cosα) = (3sinα - 3sin³α)/(3cosα - 3cos³α) = ( 3sinα•(1 - sin²α) )/( 3cosα•(1 - cos²α) ) = (sinα•cos²α)/(cosα•sin²α) = cosα/sinα = ctgα