The minimum value of \[{{2}^{\sin x}}+{{2}^{\cos x}}\]is:
(a) \[{{2}^{1-\dfrac{1}{\sqrt{2}}}}\]
(b) \[{{2}^{1+\dfrac{1}{\sqrt{2}}}}\]
(c) \[{{2}^{\sqrt{2}}}\]
(d) \[2\]

Hint: Use Arithmetic mean\[\left( AM \right)\]\[\ge \]Geometric mean\[\left( GM \right)\]between \[{{2}^{\sin x}}\]and \[{{2}^{\cos x}}\].

Here we have to find the minimum value of \[{{2}^{\sin x}}+{{2}^{\cos x}}\].
We know that
Arithmetic mean \[\ge \] Geometric mean
Or, \[AM\ge GM....\left( i \right)\]
For any two values, say \[a\]and \[b\],
\[AM=\dfrac{a+b}{2}\]
And \[GM=\sqrt{ab}\]
Considering \[a={{2}^{\sin x}}\]and \[b={{2}^{\cos x}}\]
We get \[AM=\dfrac{{{2}^{\sin x}}+{{2}^{\cos x}}}{2}\]
And \[GM=\sqrt{{{2}^{\sin x}}{{.2}^{\cos x}}}\]
Also, \[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
Therefore, \[GM=\sqrt{{{2}^{\sin x+\cos x}}}\]
By putting value of \[AM\]and \[GM\]in equation \[\left( i \right)\]
We get, \[\dfrac{{{2}^{\sin x}}+{{2}^{\cos x}}}{2}\ge \sqrt{{{2}^{\sin x+\cos x}}}\]
By cross multiplying, we get
\[={{2}^{\sin x}}+{{2}^{\cos x}}\ge 2{{\left( {{2}^{\sin x+\cos x}} \right)}^{\dfrac{1}{2}}}\]
We know that minimum value of
\[a\sin x+b\cos x=-\sqrt{{{a}^{2}}+{{b}^{2}}}\]
Therefore, minimum value of
\[\sin x+\cos x=-\sqrt{{{1}^{2}}+{{1}^{2}}}=-\sqrt{2}\]
Therefore, minimum value of
\[{{2}^{\sin x+\cos x}}={{2}^{-\sqrt{2}}}\]
Hence, \[{{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1}}{{\left( {{2}^{-\sqrt{2}}} \right)}^{\dfrac{1}{2}}}\]
\[={{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1}}{{.2}^{-\dfrac{1}{\sqrt{2}}}}\]
We know that \[{{a}^{m}}.{{a}^{n}}={{a}^{m+n}}\]
Therefore, \[{{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}\]
Hence, \[{{2}^{\sin x}}+{{2}^{\cos x}}\]is always greater than or equal to \[{{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}\].
That means, the minimum value of \[{{2}^{\sin x}}+{{2}^{\cos x}}\]is \[{{2}^{\left( 1-\dfrac{1}{\sqrt{2}} \right)}}\].
Therefore, option (a) is correct.

Note: In questions involving maxima and minima in trigonometry, students must try to use the approach
 of \[AM\ge GM\] for once and not always try to solve the question only through trigonometric
 equations and functions.