If sin ( 90^ - ) + cos = 2 cos ( 90^ - ) , then the value of cosec is?

Answer: Option C

$$\begin{array}{rl}& \text{sin}\left({90}^{\circ }-\theta \right)+\text{cos}\theta =\sqrt{2}\text{cos}\left({90}^{\circ }-\theta \right)\\ & \Rightarrow \text{cos}\theta +\text{cos}\theta =\sqrt{2}\sin \theta \\ & \Rightarrow \frac{2\cos \theta }{\sin \theta }=\sqrt{2}\\ & \Rightarrow \cot \theta =\frac{1\to \text{B}}{\sqrt{2}\to \text{P}}\\ & \text{So, H}\to \text{alignment}\\ & \therefore \text{cosec}\theta =\frac{\text{H}}{\text{P}}=\frac{\sqrt{3}}{\sqrt{2}}\\ & =\sqrt{\frac{3}{2}}\end{array}$$