The numerical value of
`cot 18^(@) (cot 72^(@) cos^(2) 22^(@) + (1)/(tan 72^(@) sec^(2) 68^(@)))` is

Options

A. 1
B. `sqrt2`
C. 3
D. `1/sqrt3`

Answer

To solve the expression \( \cot 18^\circ \left( \cot 72^\circ \cos^2 22^\circ + \frac{1}{\tan 72^\circ \sec^2 68^\circ} \right) \), we will break it down step by step.

### Step 1: Rewrite cotangent and tangent
We know that:
\[ \cot 72^\circ = \frac{1}{\tan 72^\circ} \]
Thus, we can rewrite the expression:
\[ \cot 18^\circ \left( \frac{1}{\tan 72^\circ} \cos^2 22^\circ + \frac{1}{\tan 72^\circ \sec^2 68^\circ} \right) \]

### Step 2: Factor out \(\frac{1}{\tan 72^\circ}\)
Factoring out \(\frac{1}{\tan 72^\circ}\) gives us:
\[ \cot 18^\circ \cdot \frac{1}{\tan 72^\circ} \left( \cos^2 22^\circ + \frac{1}{\sec^2 68^\circ} \right) \]

### Step 3: Simplify \(\sec^2 68^\circ\)
Recall that:
\[ \sec^2 68^\circ = \frac{1}{\cos^2 68^\circ} \]
Thus:
\[ \frac{1}{\sec^2 68^\circ} = \cos^2 68^\circ \]
So we can rewrite the expression as:
\[ \cot 18^\circ \cdot \frac{1}{\tan 72^\circ} \left( \cos^2 22^\circ + \cos^2 68^\circ \right) \]

### Step 4: Use the identity for cosine
Using the identity \(\cos^2 68^\circ = \sin^2 22^\circ\) (since \(68^\circ + 22^\circ = 90^\circ\)):
\[ \cos^2 22^\circ + \cos^2 68^\circ = \cos^2 22^\circ + \sin^2 22^\circ = 1 \]

### Step 5: Substitute back into the expression
Now we substitute back:
\[ \cot 18^\circ \cdot \frac{1}{\tan 72^\circ} \cdot 1 = \cot 18^\circ \cdot \frac{1}{\tan 72^\circ} \]

### Step 6: Simplify \(\cot 18^\circ\) and \(\tan 72^\circ\)
Using the identity \(\tan 72^\circ = \cot 18^\circ\):
\[ \cot 18^\circ \cdot \frac{1}{\tan 72^\circ} = \cot 18^\circ \cdot \frac{1}{\cot 18^\circ} = 1 \]

### Final Answer
The numerical value of the expression is:
\[ \boxed{1} \]