Evaluate : `(sec^(2)theta-cot^(2)(90^(@)-theta))/("cosec"^(2)67^(@)-tan^(2)23^(2))+(sin^(2)40^(@)+sin^(2)50^(@))`

Answer

To evaluate the expression

\[ \frac{\sec^2 \theta - \cot^2 (90^\circ - \theta)}{\csc^2 67^\circ - \tan^2 23^\circ} + (\sin^2 40^\circ + \sin^2 50^\circ), \]

we will simplify it step by step.

### Step 1: Simplify the numerator

The numerator is

\[ \sec^2 \theta - \cot^2 (90^\circ - \theta). \]

Using the identity \(\cot(90^\circ - \theta) = \tan \theta\), we can rewrite this as:

\[ \sec^2 \theta - \tan^2 \theta. \]

### Step 2: Use the identity

We know from trigonometric identities that:

\[ \sec^2 \theta - \tan^2 \theta = 1. \]

So, the numerator simplifies to:

\[ 1. \]

### Step 3: Simplify the denominator

The denominator is

\[ \csc^2 67^\circ - \tan^2 23^\circ. \]

Using the identity \(\csc(90^\circ - \theta) = \sec \theta\), we can rewrite \(\csc^2 67^\circ\) as:

\[ \sec^2 23^\circ. \]

Thus, the denominator becomes:

\[ \sec^2 23^\circ - \tan^2 23^\circ. \]

### Step 4: Use the identity again

Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\) again, we find that:

\[ \sec^2 23^\circ - \tan^2 23^\circ = 1. \]

### Step 5: Combine the results

Now substituting back into the expression, we have:

\[ \frac{1}{1} + (\sin^2 40^\circ + \sin^2 50^\circ). \]

### Step 6: Simplify \(\sin^2 40^\circ + \sin^2 50^\circ\)

Using the identity \(\sin(90^\circ - \theta) = \cos \theta\), we can write:

\[ \sin^2 50^\circ = \cos^2 40^\circ. \]

Thus,

\[ \sin^2 40^\circ + \sin^2 50^\circ = \sin^2 40^\circ + \cos^2 40^\circ = 1. \]

### Step 7: Final calculation

Now we can combine everything:

\[ 1 + 1 = 2. \]

Thus, the final answer is

\[ \boxed{2}. \]