Prove that sinB/1-cosB -1+cosB/sinB =0

To make the subtraction between the two fractions, they have to have the same denominator. For this reason, we'll do the amplification of the first fraction, with the denominator of the second, namely sinB, and we'll do the amplification of the second fraction with the denominator of the first one, namely (1-cosB).

The expression will appear in this way:

 [(sinB*sinB)-(1-cosB)*(1+cosB)] : [sinB*(1-cosB)]=0

Also, doing the cross muliplying between the left terms of the expression and the right terms, we'll write, even more simple:

 [(sinB*sinB)-(1-cosB)*(1+cosB)]=[sinB*(1-cosB)]*0

 [(sinB*sinB)-(1-cosB)*(1+cosB)]=

Now, all we have to do is to open the brackets and solve the expression:

sinB*sinB= (sinB)^2

(1-cosB)*(1+cosB)=1-(cos B)^2

(sinB)^2-[1-(cos B)^2]=0

(sinB)^2-1+(cos B)^2=0

We note that in the expression above, are added the squares of the joint functions, (SIN B)^2 and (COS B)^2,of the same angle B.

(sinB)^2 + (cos B)^2=1

We'll substitute the adding with sum, in the expression above:

(sinB)^2-1+(cos B)^2=0 => 1-1=0 => 0=0 TRUE