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Answer variants:
1+sinθ21sin2θ
1+sinθcosθ
1+sinθ2cos2θ
1+sinθ21+sin2θ
1cosθ+sinθcosθ
1+sinθ1sinθ×1+sinθ1+sinθ
1+sinθ1sinθ×1sinθ1sinθ
Verify that 1+sinθ1sinθ  \(=\) secθ+tanθ.
 
Proof:
 
LHS \(=\) 1+sinθ1sinθ
 
\(=\)
 
\(=\)
 
\(=\)
 
\(=\)
 
\(=\)
 
\(=\) secθ+tanθ \(=\) RHS