Here is how you can prove the identity is true. First, express all terms as either sin(x)sin(x) or cos(x)cos(x)
(1/sin(x)+1/cos(x))/(sin(x)+cos(x))=cos(x)/sin(x)+sin(x)/cos(x)1sin(x)+1cos(x)sin(x)+cos(x)=cos(x)sin(x)+sin(x)cos(x)
Then multiply both sides by sin(x)cos(x)sin(x)cos(x)
sin(x)cos(x)xx(1/sin(x)+1/cos(x))/(sin(x)+cos(x))sin(x)cos(x)×1sin(x)+1cos(x)sin(x)+cos(x)
=sin(x)cos(x)xx(cos(x)/sin(x)+sin(x)/cos(x))=sin(x)cos(x)×(cos(x)sin(x)+sin(x)cos(x))
The left hand side multiplies sin(x)cos(x)sin(x)cos(x) through the numerator, giving
(cos(x)+sin(x))/(sin(x)+cos(x))=1cos(x)+sin(x)sin(x)+cos(x)=1
The right hand side multiplies sin(x)cos(x)sin(x)cos(x) though each term, giving
=cos^2(x)+sin^2(x)=1=cos2(x)+sin2(x)=1
Because the left hand side equals the right hand side, the identity is proven.
