Find the value of `(tan 9^(@)+cot9^(@))/(tan27^(@)+cot27^(@))`.

Explanation

To solve the problem \((\tan 9^\circ + \cot 9^\circ) / (\tan 27^\circ + \cot 27^\circ)\), we can follow these steps:

### Step 1: Rewrite cotangent in terms of sine and cosine
We know that \(\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\). Therefore, we can rewrite the expression as:

\[ \frac{\tan 9^\circ + \cot 9^\circ}{\tan 27^\circ + \cot 27^\circ} = \frac{\tan 9^\circ + \frac{\cos 9^\circ}{\sin 9^\circ}}{\tan 27^\circ + \frac{\cos 27^\circ}{\sin 27^\circ} \]

### Step 2: Combine the terms in the numerator and denominator
Combining the terms in the numerator and denominator gives us:

\[ = \frac{\frac{\sin 9^\circ}{\cos 9^\circ} + \frac{\cos 9^\circ}{\sin 9^\circ}}{\frac{\sin 27^\circ}{\cos 27^\circ} + \frac{\cos 27^\circ}{\sin 27^\circ}} \]

### Step 3: Find a common denominator
The common denominator for both the numerator and denominator is \(\sin 9^\circ \cos 9^\circ\) and \(\sin 27^\circ \cos 27^\circ\) respectively:

\[ = \frac{\frac{\sin^2 9^\circ + \cos^2 9^\circ}{\sin 9^\circ \cos 9^\circ}}{\frac{\sin^2 27^\circ + \cos^2 27^\circ}{\sin 27^\circ \cos 27^\circ}} \]

### Step 4: Simplify using the Pythagorean identity
Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ = \frac{\frac{1}{\sin 9^\circ \cos 9^\circ}}{\frac{1}{\sin 27^\circ \cos 27^\circ}} = \frac{\sin 27^\circ \cos 27^\circ}{\sin 9^\circ \cos 9^\circ} \]

### Step 5: Use the double angle identity
We can express \(\sin \theta \cos \theta\) using the double angle identity:

\[ = \frac{\frac{1}{2} \sin 54^\circ}{\frac{1}{2} \sin 18^\circ} = \frac{\sin 54^\circ}{\sin 18^\circ} \]

### Step 6: Simplify using sine properties
Using the property that \(\sin(90^\circ - \theta) = \cos(\theta)\):

\[ = \frac{\cos 36^\circ}{\sin 18^\circ} \]

### Step 7: Substitute known values
We know that \(\cos 36^\circ = \frac{\sqrt{5} + 1}{4}\) and \(\sin 18^\circ = \frac{\sqrt{5} - 1}{4}\):

\[ = \frac{\frac{\sqrt{5} + 1}{4}}{\frac{\sqrt{5} - 1}{4}} = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \]

### Step 8: Rationalize the denominator
To rationalize the denominator:

\[ = \frac{(\sqrt{5} + 1)(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)} = \frac{5 + 2\sqrt{5} + 1}{5 - 1} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]

### Final Answer
Thus, the value of \(\frac{\tan 9^\circ + \cot 9^\circ}{\tan 27^\circ + \cot 27^\circ}\) is:

\[ \boxed{\frac{3 + \sqrt{5}}{2}} \]