If $$\sin(90^\circ-\theta)\ +\ cos\ \theta = \sqrt{2}\ cos(90^\circ-\theta)$$, then the value of cosec $$\theta$$ is

Given,

$$\sin(90^\circ-\theta)\ +\ cos\ \theta = \sqrt{2}\ cos(90^\circ-\theta)$$.............(1)

We know that, $$\sin(90^{\circ}-\theta)$$ = $$\cos\theta$$ and $$\cos(90^{\circ}-\theta)$$ = $$\sin\theta$$

Now, Equation (1) can be written as,

$$\cos\theta\ +\ cos\ \theta = \sqrt{2}\ sin\theta$$ (or) $$2 \cos\theta\ = \sqrt{2}\ sin\theta$$

Squaring on both sides we get,

$$2\cos^{2}\theta\ =\ sin^{2}\theta$$

$$\Rightarrow 2(1 - sin^{2}\theta) = sin^{2}\theta \Rightarrow 2 - 2sin^{2}\theta = sin^{2}\theta$$

$$\Rightarrow 2 = 3sin^{2}\theta \Rightarrow 2 = 3(\frac{1}{cosec^{2}\theta}) \Rightarrow cosec^{2}\theta = \frac{3}{2}$$

$$\Rightarrow cosec\theta = \sqrt{\frac{3}{2}}$$

Hence, option B is the correct answer.