The value of `cot18^(@)(cot72^(@)cos^(2)22^(@)+(1)/(tan72^(@)sec^(2)68^(@)))` is

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 the expression
We start with the expression:
\[ \cot 18^\circ \left( \cot 72^\circ \cos^2 22^\circ + \frac{1}{\tan 72^\circ \sec^2 68^\circ} \right) \]

### Step 2: Simplify the second term
Recall that:
- \( \tan \theta = \frac{1}{\cot \theta} \)
- \( \sec^2 \theta = \frac{1}{\cos^2 \theta} \)

Thus, we can rewrite the second term:
\[ \frac{1}{\tan 72^\circ \sec^2 68^\circ} = \frac{1}{\tan 72^\circ} \cdot \cos^2 68^\circ \]
Since \( \sec^2 68^\circ = \frac{1}{\cos^2 68^\circ} \).

### Step 3: Substitute and combine
Now, substituting back into the expression, we have:
\[ \cot 18^\circ \left( \cot 72^\circ \cos^2 22^\circ + \cot 72^\circ \cos^2 68^\circ \right) \]
Factoring out \( \cot 72^\circ \):
\[ \cot 18^\circ \cot 72^\circ \left( \cos^2 22^\circ + \cos^2 68^\circ \right) \]

### Step 4: Use the identity for cosine
Using the identity \( \cos(90^\circ - \theta) = \sin \theta \):
\[ \cos^2 68^\circ = \cos^2(90^\circ - 22^\circ) = \sin^2 22^\circ \]
Thus, we can rewrite:
\[ \cos^2 22^\circ + \cos^2 68^\circ = \cos^2 22^\circ + \sin^2 22^\circ = 1 \]

### Step 5: Substitute back into the expression
Now substituting this back, we get:
\[ \cot 18^\circ \cot 72^\circ \cdot 1 = \cot 18^\circ \cot 72^\circ \]

### Step 6: Use the cotangent identity
Recall that:
\[ \cot 72^\circ = \tan(90^\circ - 72^\circ) = \tan 18^\circ \]
Thus:
\[ \cot 18^\circ \cot 72^\circ = \cot 18^\circ \tan 18^\circ \]
Since \( \cot \theta = \frac{1}{\tan \theta} \), we have:
\[ \cot 18^\circ \tan 18^\circ = 1 \]

### Final Result
Thus, the value of the original expression is:
\[ \boxed{1} \]