If 1 + 2tan²θ + 2sinθsec²θ = $$\frac{a}{b}$$, 0° < θ < 90°, then $$\frac{{a + b}}{{a - b}}?$$

Solution (By Examveda Team)

$$\eqalign{ & 1 + 2{\tan ^2}\theta + 2\sin \theta {\sec ^2}\theta = \frac{a}{b} \cr & 1 + {\tan ^2}\theta + {\tan ^2}\theta + 2\sin \theta \frac{1}{{{{\cos }^2}\theta }} = \frac{a}{b} \cr & {\sec ^2}\theta + {\tan ^2}\theta + 2\tan \theta \sec \theta = \frac{a}{b} \cr & {\left( {\sec \theta + \tan \theta } \right)^2} = \frac{a}{b} \cr & {\left( {\frac{1}{{\cos \theta }} + \frac{{\sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & {\left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)^2} = \frac{a}{b} \cr & \frac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 + {{\sin }^2}\theta }} = \frac{a}{b} \cr & \frac{{1 + \sin \theta }}{{1 - \sin \theta }} = \frac{a}{b} \cr & {\text{Use componendo and dividendo}} \cr & \frac{1}{{\sin \theta }} = {\text{cosec}}\,\theta = \frac{{a + b}}{{a - b}} \cr} $$