Given :

\(\frac{(1 + tanθ + secθ)(1 + cotθ - cosecθ)}{(secθ + tanθ)(1 - sinθ)}\)

Concept used :

tanθ = sinθ/cosθ

secθ = 1/cosθ 

cotθ = cosθ\sinθ

Solution :

 \(\frac{(1 + tanθ + secθ)(1 + cotθ - cosecθ)}{(secθ + tanθ)(1 - sinθ)}\)

\( = \;\frac{{\left( {1 + \frac{{sinθ }}{{cosθ }} + \frac{1}{{cosθ }}} \right)\left( {1 + \frac{{cosθ }}{{sinθ }} - \frac{1}{{sinθ }}} \right)}}{{\left( {\frac{1}{{cosθ }} + \frac{{sinθ }}{{cosθ }}\;} \right)\left( {1 - sinθ } \right)}}\)

\( = \;\frac{{\left( {\frac{1}{{sinθ cosθ }}} \right)\left( {cosθ + sinθ + 1} \right)\left( {sinθ + cosθ - 1} \right)}}{{\frac{1}{{cosθ }}\left( {1 + sinθ } \right)\left( {1 - sinθ } \right)}}\)

\( = \frac{1}{{sinθ }}\;\;\frac{{\left[ {{{\left( {cosθ + sinθ } \right)}^2} - {1^2}} \right]}}{{\left( {1 - {{\sin }^2}θ } \right)}}\)

\( = \frac{1}{{sinθ }}\;\frac{{\left( {{{\cos }^2}θ + {{\sin }^2}θ + 2sinθ cosθ } \right) - 1}}{{\left( {1 - {{\sin }^2}θ } \right)}}\)

\( = \frac{1}{{sinθ }}\;\frac{{2sinθ cosθ }}{{{{\cos }^2}θ }}\)

= 2/cosθ 

= 2secθ 

∴ the required value is 2secθ .