Question

已知(tgA+1)/(1-tgA)=3+2根号2 求sinA和cosA

Answer 1

(tgA+1)/(1-tgA)=3+2根号2

=>tgA+1=(3+2根号2)*(1-tgA)

=>tgA+1=3-3tgA+2根号2-2根号2tgA

=>2tgA+2根号2tgA=2+2根号2

=>(2+2根号2)tgA=2+2根号2

=>tgA=1

即sinA=cosA

sinA=cosA=(根号2)/2