ABCD is a rectangle of which AC is a diagonal .The value of `(tan^(2)angleCAD+1)sin^(2)angleBAC` is

Options

A. 2
B. `(1)/(4)`
C. 1
D. 0

Answer

To solve the problem, we need to find the value of \( \left( \tan^2 \angle CAD + 1 \right) \sin^2 \angle BAC \) in the rectangle ABCD where AC is a diagonal.

### Step-by-step Solution:

1. **Identify the Angles**:
- In rectangle ABCD, the diagonal AC divides the rectangle into two right triangles: \( \triangle ABC \) and \( \triangle ADC \).
- Since ABCD is a rectangle, the angles \( \angle CAD \) and \( \angle BAC \) are both \( 45^\circ \) because the diagonal bisects the right angle at A.

2. **Calculate \( \tan^2 \angle CAD \)**:
- We know that \( \angle CAD = 45^\circ \).
- Therefore, \( \tan 45^\circ = 1 \).
- Thus, \( \tan^2 \angle CAD = 1^2 = 1 \).

3. **Calculate \( \sin^2 \angle BAC \)**:
- Again, since \( \angle BAC = 45^\circ \), we have \( \sin 45^\circ = \frac{1}{\sqrt{2}} \).
- Therefore, \( \sin^2 \angle BAC = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} \).

4. **Substitute Values into the Expression**:
- Now substitute the values into the expression:
\[ \left( \tan^2 \angle CAD + 1 \right) \sin^2 \angle BAC = \left( 1 + 1 \right) \cdot \frac{1}{2} \]
- Simplifying this gives:
\[ 2 \cdot \frac{1}{2} = 1 \]

5. **Final Answer**:
- Therefore, the value of \( \left( \tan^2 \angle CAD + 1 \right) \sin^2 \angle BAC \) is \( 1 \).

### Summary:
The final answer is \( 1 \).