Evaluate: `(sec 29^(@))/("cosec" 61^(@)) + 2 cot 8^(@) cot 17^(@) cot 45^(@) cot 73^(@) cot82^(@) - 3(sin^(2) 38^(@) + sin^(2) 52^(@)) + 4[cos theta sin(90^(@) - theta) + sin theta cos (90^(@) - theta)]`

Answer

To evaluate the expression

\[ \frac{\sec 29^\circ}{\csc 61^\circ} + 2 \cot 8^\circ \cot 17^\circ \cot 45^\circ \cot 73^\circ \cot 82^\circ - 3(\sin^2 38^\circ + \sin^2 52^\circ) + 4\left[\cos \theta \sin(90^\circ - \theta) + \sin \theta \cos(90^\circ - \theta)\right], \]

we will break it down step by step.

### Step 1: Evaluate \(\frac{\sec 29^\circ}{\csc 61^\circ}\)

Using the identity \(\csc \theta = \sec(90^\circ - \theta)\):

\[ \csc 61^\circ = \sec(90^\circ - 61^\circ) = \sec 29^\circ. \]

Thus,

\[ \frac{\sec 29^\circ}{\csc 61^\circ} = \frac{\sec 29^\circ}{\sec 29^\circ} = 1. \]

### Step 2: Evaluate \(2 \cot 8^\circ \cot 17^\circ \cot 45^\circ \cot 73^\circ \cot 82^\circ\)

Using the identity \(\cot(90^\circ - \theta) = \tan \theta\):

\[ \cot 73^\circ = \tan 17^\circ \quad \text{and} \quad \cot 82^\circ = \tan 8^\circ. \]

Thus, we can rewrite:

\[ \cot 8^\circ \cot 17^\circ \cot 45^\circ \cot 73^\circ \cot 82^\circ = \cot 8^\circ \tan 8^\circ \cot 17^\circ \tan 17^\circ \cdot 1. \]

Since \(\cot \theta \tan \theta = 1\):

\[ \cot 8^\circ \tan 8^\circ = 1 \quad \text{and} \quad \cot 17^\circ \tan 17^\circ = 1. \]

Thus,

\[ \cot 8^\circ \cot 17^\circ \cot 45^\circ \cot 73^\circ \cot 82^\circ = 1. \]

So,

\[ 2 \cot 8^\circ \cot 17^\circ \cot 45^\circ \cot 73^\circ \cot 82^\circ = 2. \]

### Step 3: Evaluate \(-3(\sin^2 38^\circ + \sin^2 52^\circ)\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \sin^2 52^\circ = \cos^2 38^\circ. \]

Thus,

\[ \sin^2 38^\circ + \sin^2 52^\circ = \sin^2 38^\circ + \cos^2 38^\circ = 1. \]

So,

\[ -3(\sin^2 38^\circ + \sin^2 52^\circ) = -3 \cdot 1 = -3. \]

### Step 4: Evaluate \(4\left[\cos \theta \sin(90^\circ - \theta) + \sin \theta \cos(90^\circ - \theta)\right]\)

Using the identity \(\sin(90^\circ - \theta) = \cos \theta\):

\[ \cos \theta \sin(90^\circ - \theta) + \sin \theta \cos(90^\circ - \theta) = \cos \theta \cos \theta + \sin \theta \sin \theta = \cos^2 \theta + \sin^2 \theta = 1. \]

Thus,

\[ 4\left[\cos \theta \sin(90^\circ - \theta) + \sin \theta \cos(90^\circ - \theta)\right] = 4 \cdot 1 = 4. \]

### Final Calculation

Now, we combine all parts:

\[ 1 + 2 - 3 + 4 = 4. \]

Thus, the final answer is

\[ \boxed{4}. \]