If $$\frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7},\,\theta ,$$     lies in first quadrant, then the value of $$\frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }}$$   is:

Solution(By Examveda Team)

$$\eqalign{ & \frac{{\sec \theta - \tan \theta }}{{\sec \theta + \tan \theta }} = \frac{1}{7} \cr & 7\sec \theta - 7\tan \theta = \sec \theta + \tan \theta \cr & 6\sec \theta = 8\tan \theta \cr & \sin \theta = \frac{6}{8} = \frac{3}{4} = \frac{P}{H} \cr & B = \sqrt {{4^2} - {3^2}} = \sqrt 7 \cr & \Rightarrow \frac{{{\text{cosec}}\,\theta + {{\cot }^2}\theta }}{{{\text{cosec}}\,\theta - {{\cot }^2}\theta }} \cr & = \frac{{\frac{H}{P} + {{\left( {\frac{B}{P}} \right)}^2}}}{{\frac{H}{P} - {{\left( {\frac{B}{P}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}}{{\frac{4}{3} - {{\left( {\frac{{\sqrt 7 }}{3}} \right)}^2}}} \cr & = \frac{{\frac{4}{3} + \frac{7}{9}}}{{\frac{4}{3} - \frac{7}{9}}} \cr & = \frac{{\frac{{12 + 7}}{9}}}{{\frac{{12 - 7}}{9}}} \cr & = \frac{{19}}{5} \cr} $$