If 1 + 2tan2θ + 2sinθsec2θ = a/b , 0° < θ < 90°, then a + ba - b?

Answer: Option B

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