Evaluate $\frac{cos 45^\circ }{tan 30^\circ + sin 60^\circ}$

Evaluate $\displaystyle \frac{cos 45^\circ }{tan 30^\circ + sin 60^\circ}$

We need to evaluate the expression:

$ \displaystyle \frac{\cos 45^\circ }{\tan 30^\circ + \sin 60^\circ} $

### Step 1: Substitute Trigonometric Values
From trigonometric ratios:

$ \cos 45^\circ = \frac{1}{\sqrt{2}} $
$ \tan 30^\circ = \frac{1}{\sqrt{3}} $
$ \sin 60^\circ = \frac{\sqrt{3}}{2} $

Substituting these values:

$ \displaystyle \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2}} $

### Step 2: Simplify the Denominator
Taking LCM in the denominator:

$ \displaystyle \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} $

Thus, the expression simplifies to:

$ \displaystyle \frac{\frac{1}{\sqrt{2}}}{\frac{5}{2\sqrt{3}}} $

### Step 3: Multiply by the Reciprocal
Rewriting:

$ \displaystyle \frac{1}{\sqrt{2}} \times \frac{2\sqrt{3}}{5} = \frac{2\sqrt{3}}{5\sqrt{2}} $

Rationalizing the denominator:

$ \displaystyle \frac{2\sqrt{3} \times \sqrt{2}}{5\sqrt{2} \times \sqrt{2}} = \frac{2\sqrt{6}}{10} $

$ \displaystyle = \frac{\sqrt{6}}{5} $

### Final Answer:
$ \displaystyle \frac{\sqrt{6}}{5} $

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