Find the value of `sec^(2)10^(@)-cot^(2)80^(@)+(sin15^(@)cos75^(@)+cos15^(@)sin75^(@))/(costhetasin(90^(@)-theta)+sinthetacos(90^(@)-theta))`.

Answer

To solve the expression

\[ \sec^2(10^\circ) - \cot^2(80^\circ) + \frac{\sin(15^\circ)\cos(75^\circ) + \cos(15^\circ)\sin(75^\circ)}{\cos(\theta)\sin(90^\circ - \theta) + \sin(\theta)\cos(90^\circ - \theta)} \]

we will break it down step by step.

### Step 1: Simplify \(\sec^2(10^\circ) - \cot^2(80^\circ)\)

We know that:

\[ \sec^2(\theta) = \frac{1}{\cos^2(\theta)} \]
and
\[ \cot^2(\theta) = \frac{\cos^2(\theta)}{\sin^2(\theta)} \]

Thus, we can rewrite \(\sec^2(10^\circ)\) and \(\cot^2(80^\circ)\):

\[ \sec^2(10^\circ) = \frac{1}{\cos^2(10^\circ)} \]
\[ \cot^2(80^\circ) = \frac{\cos^2(80^\circ)}{\sin^2(80^\circ)} \]

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

\[ \cot(80^\circ) = \tan(10^\circ) \implies \cot^2(80^\circ) = \tan^2(10^\circ) = \frac{\sin^2(10^\circ)}{\cos^2(10^\circ)} \]

Now we can rewrite the expression:

\[ \sec^2(10^\circ) - \cot^2(80^\circ) = \frac{1}{\cos^2(10^\circ)} - \frac{\sin^2(10^\circ)}{\cos^2(10^\circ)} = \frac{1 - \sin^2(10^\circ)}{\cos^2(10^\circ)} \]

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

\[ \sec^2(10^\circ) - \cot^2(80^\circ) = \frac{\cos^2(10^\circ)}{\cos^2(10^\circ)} = 1 \]

### Step 2: Simplify the numerator of the fraction

The numerator is:

\[ \sin(15^\circ)\cos(75^\circ) + \cos(15^\circ)\sin(75^\circ) \]

Using the sine addition formula \(\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b)\):

\[ \sin(15^\circ + 75^\circ) = \sin(90^\circ) = 1 \]

### Step 3: Simplify the denominator of the fraction

The denominator is:

\[ \cos(\theta)\sin(90^\circ - \theta) + \sin(\theta)\cos(90^\circ - \theta) \]

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

\[ \cos(\theta)\cos(\theta) + \sin(\theta)\sin(\theta) = \cos^2(\theta) + \sin^2(\theta) = 1 \]

### Step 4: Combine all parts

Putting it all together:

\[ \sec^2(10^\circ) - \cot^2(80^\circ) + \frac{1}{1} = 1 + 1 = 2 \]

### Final Answer

The value of the expression is:

\[ \boxed{2} \]