Question

sinα+cosα=4/3求tanα+1/tanα=?

Answer 1

设含α的直角三角形边长为a,b,c

sinα=a/c

cosα=b/c

(a+b)/c=4/3 c=3*(a+b)/4 c^2=9/16*(a+b)^2

a^2+b^2=c^2结合上面

得7*(a^2+b^2)=18ab

tanα+1/tanα=a/b+b/a=(a^2+b^2)/ab=18/7