In a triangle `angleB=90^(@)` and `angleA=30^(@)` then `Sin^(2)A-Cos^(2)A+Tan^(2)A` :

Options

A. `(-3)/(12)`
B. `1/6`
C. `2/6`
D. `(-1)/(6)`

Answer

To solve the problem, we need to calculate the expression \( \sin^2 A - \cos^2 A + \tan^2 A \) for the given angles in the triangle.

### Step-by-Step Solution:

1. **Identify the angles**: We know that in the triangle, \( \angle B = 90^\circ \) and \( \angle A = 30^\circ \). Therefore, \( \angle C = 90^\circ - 30^\circ = 60^\circ \).

2. **Calculate \( \sin^2 A \)**:
\[ \sin A = \sin 30^\circ = \frac{1}{2} \]
\[ \sin^2 A = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \]

3. **Calculate \( \cos^2 A \)**:
\[ \cos A = \cos 30^\circ = \frac{\sqrt{3}}{2} \]
\[ \cos^2 A = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \]

4. **Calculate \( \tan^2 A \)**:
\[ \tan A = \tan 30^\circ = \frac{1}{\sqrt{3}} \]
\[ \tan^2 A = \left( \frac{1}{\sqrt{3}} \right)^2 = \frac{1}{3} \]

5. **Substitute values into the expression**:
\[ \sin^2 A - \cos^2 A + \tan^2 A = \frac{1}{4} - \frac{3}{4} + \frac{1}{3} \]

6. **Combine the fractions**:
- The common denominator for \( \frac{1}{4} \), \( \frac{3}{4} \), and \( \frac{1}{3} \) is 12.
- Convert each fraction:
\[ \frac{1}{4} = \frac{3}{12}, \quad \frac{3}{4} = \frac{9}{12}, \quad \frac{1}{3} = \frac{4}{12} \]
- Substitute back:
\[ \frac{3}{12} - \frac{9}{12} + \frac{4}{12} = \frac{3 - 9 + 4}{12} = \frac{-2}{12} \]

7. **Simplify the result**:
\[ \frac{-2}{12} = \frac{-1}{6} \]

### Final Answer:
Thus, the value of \( \sin^2 A - \cos^2 A + \tan^2 A \) is \( \frac{-1}{6} \).