1708\hspace{-16}$Given $\bf{2^{\sin x}+2^{\cos x}>1}$\\\\\\ Now we know that $\bf{-1\leq \sin x\;,\cos x\leq 1\;,\forall\;x\in \mathbb{R}}$\\\\\\ So $\bf{2^{-1}\leq 2^{\sin x}\;,2^{\cos x}\leq 2\;\;,\forall\; x\in \mathbb{R}}$\\\\\\ So $\bf{2^{\sin x}+2^{\cos x}\geq \frac{1}{2}+\frac{1}{2}=1}$\\\\\\ So Here Equality hold , When $\bf{2^{\sin x}=\frac{1}{2}=2^{-1}}$ and $\bf{2^{\cos x}=\frac{1}{2}=2^{-1}}$\\\\\\ So $\bf{\sin x= -1}$ and $\bf{\cos x= -1}$\\\\\\ Which is never Possible for common values of $\bf{x}$\\\\\\ So $\bf{2^{\sin x}+2^{\cos x}>1\;\;,\forall\;x\in \mathbb{R}}$