Prove that:
 
sinθ1+cosθ+1+cosθsinθ=2cosecθ
 
 
 Proof
L.H.S \(=\) sinθ1+cosθ+1+cosθsinθ
 
 
\(LHS=\)
 
By applying \((a+b)^2\) identity, then we get,
 
\(LHS=\)
 
By applying the known identity then we get, 
 
 
By simplyfing this then we get,
 
\(= 2 cosec \theta\)
 
\(=\) R.H.S
 
Hence proved.
Answer variants:
=sin2θ+1+cosθ2sinθ1+cosθ
=2+2cosθsinθ1+cosθ
=sin2θ+cos2θ+12cosθsinθ1+cosθ
=sin2θ+cos2θ+1+2cosθsinθ1+cosθ
=2+cosθsinθ1+cosθ