The value of `[(sin^2 24^@+ sin^2 66^@)/(cos^2 24^@+cos^2 66^@)+ sin^2 61^@+cos 61^@ sin 29^@]` is:

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To solve the given expression:

\[ \frac{\sin^2 24^\circ + \sin^2 66^\circ}{\cos^2 24^\circ + \cos^2 66^\circ} + \sin^2 61^\circ + \cos 61^\circ \sin 29^\circ \]

we will break it down step by step.

### Step 1: Simplify \(\sin^2 66^\circ\) and \(\cos^2 66^\circ\)

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

\[ \sin^2 66^\circ = \sin^2(90^\circ - 24^\circ) = \cos^2 24^\circ \]
\[ \cos^2 66^\circ = \cos^2(90^\circ - 24^\circ) = \sin^2 24^\circ \]

### Step 2: Substitute back into the expression

Now, substituting these identities into the expression:

\[ \frac{\sin^2 24^\circ + \cos^2 24^\circ}{\cos^2 24^\circ + \sin^2 24^\circ} \]

### Step 3: Simplify the fraction

Since \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \frac{1}{1} = 1 \]

### Step 4: Simplify \(\sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\)

Next, we need to simplify \(\sin^2 61^\circ + \cos 61^\circ \sin 29^\circ\).

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

\[ \sin 29^\circ = \sin(90^\circ - 61^\circ) = \cos 61^\circ \]

Thus, we can rewrite \(\cos 61^\circ \sin 29^\circ\) as:

\[ \cos 61^\circ \cos 61^\circ = \cos^2 61^\circ \]

So now we have:

\[ \sin^2 61^\circ + \cos^2 61^\circ \]

### Step 5: Apply the Pythagorean identity

Again, using the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\[ \sin^2 61^\circ + \cos^2 61^\circ = 1 \]

### Step 6: Combine the results

Now, we combine the results from Step 3 and Step 5:

\[ 1 + 1 = 2 \]

### Final Answer

Thus, the value of the given expression is:

\[ \boxed{2} \]