If $$\sin Q + \cos Q = \surd2 \sin (90^\circ - Q)$$ then $$\cot Q$$ is equal to:

Given,  $$\sin Q + \cos Q = \surd2 \sin (90^\circ - Q)$$

$$=$$> $$\sin Q+\cos Q=\surd2\cos Q$$

$$=$$>  $$\sin Q=\surd2\cos Q-\cos Q$$

$$=$$>  $$\sin Q=\left(\surd2-1\right)\cos Q$$

$$=$$>  $$\frac{\cos Q}{\sin Q}=\frac{1}{\surd2-1}$$

$$=$$>  $$\cot Q=\frac{1}{\surd2-1}\times\frac{\surd2+1}{\surd2+1}$$

$$=$$>  $$\cot Q=\frac{\surd2+1}{2-1}$$

$$=$$>  $$\cot Q=\surd2+1$$

Hence, the correct answer is Option C