Correct Answer - Option 4 : 2secθ
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 .
