Assertion(A)$2^{\sin \theta }+2^{\cos \theta }\ge 2^{1-\frac{1}{\sqrt{2}}}$ for all values of $\theta$

Reason(R): The minimum value of $\sin \theta +\cos \theta \mathrm{is} -\sqrt{2}$

$$\begin{gathered} \text{We have }\boxed{\text{A.M≥G.M}} \\ \Rightarrow \frac{2^{\sin \theta }+2^{\cos \theta }}{2}\ge \sqrt{2^{\sin \theta }\cdot 2^{\cos \theta }} \\ \Rightarrow 2^{\sin \theta }+2^{\cos \theta }\ge {2.2}^{(\sin \theta +\cos \theta )/2} \\ \sin \theta +\cos \theta =\sqrt{2}\sin (\theta +\pi /4)\ge -\sqrt{2}\text{ for all real }\theta \\ \therefore 2^{\sin \theta }+2^{\cos \theta }\ge {2.2}^{(\sin \theta +\cos \theta )/2}>{2.2}^{-\sqrt{2}/2}=2^{1-(1/\sqrt{2})} \\ \boxed{2^{\sin \theta }+2^{\cos \theta }\ge 2^{1-(1/\sqrt{2})}} \end{gathered}$$