If cosθ = $$\frac{{12}}{{13}},$$ then the value of $$\frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}}$$    is:

Solution(By Examveda Team)

$$\eqalign{ & \cos \theta = \frac{{12}}{{13}} = \frac{B}{H} \cr & P = \sqrt {{{13}^2} - {{12}^2}} = 5\,{\text{cm}} \cr & \frac{{\sin \theta \left( {1 - \tan \theta } \right)}}{{\tan \theta \left( {1 + {\text{cosec}}\theta } \right)}} \cr & = \frac{{\frac{P}{H}\left( {1 - \frac{P}{B}} \right)}}{{\frac{P}{B}\left( {1 + \frac{H}{P}} \right)}} \cr & = \frac{{\frac{5}{{13}}\left( {1 - \frac{5}{{12}}} \right)}}{{\frac{5}{{12}}\left( {1 + \frac{{13}}{5}} \right)}} \cr & = \frac{{\frac{5}{{13}} \times \frac{7}{{12}}}}{{\frac{5}{{12}} \times \frac{{18}}{5}}} \cr & = \frac{{\frac{{35}}{{156}}}}{{\frac{3}{2}}} \cr & = \frac{{35 \times 2}}{{156 \times 3}} \cr & = \frac{{35}}{{234}} \cr} $$