The value of 1 + + - , if 0° < θ < 90°, is equal to:

Answer: Option D

$$\begin{array}{rl}& 1+\sqrt{\frac{\cot \theta +\cos \theta }{\cot \theta -\cos \theta }}\\ & =1+\sqrt{\frac{\frac{\cos \theta }{\sin \theta }+\cos \theta }{\frac{\cos \theta }{\sin \theta }-\cos \theta }}\\ & =1+\sqrt{\frac{\cot \theta +\sin \theta \cos \theta }{\cot \theta -\cos \theta \sin \theta }}\\ & =1+\sqrt{\frac{\cos \theta \left(1+\sin \theta \right)}{\cos \theta \left(1-\sin \theta \right)}}\\ & =1+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}\\ & \text{Rationalise,}\\ & =1+\sqrt{\frac{1+\sin \theta }{1-\sin \theta }\times \frac{1+\sin \theta }{1+\sin \theta }}\\ & =1+\sqrt{\frac{{\left(1+\sin \theta \right)}^2}{1-{\sin }^2\theta }}\\ & =1+\sqrt{{\left(1+\sin \theta \right)}^2\times {\sec }^2\theta }\\ & =1+\left(1+\sin \theta \right)\sec \theta \\ & =\sec \theta +\sin \theta \sec \theta \\ & =1+\sec \theta +\tan \theta \end{array}$$