Check that sinx1+cosx+sinx1cosx \(=\) 2cosecx.
 
Proof:
 
LHS \(=\) sinx1+cosx+sinx1cosx
 
\(=\)
 
\(=\)
 
\(=\)
 
\(=\)
 
\(=\)
 
\(=\) RHS
 
Hence, we proved.
Answer variants:
sinx1cosx+sinx1+cosx1+cosx1cosx
2sinxsin2x
2cosecx
2sinx
sinxsinxcosx+sinx+sinxcosx1cos2x