If `sin3A=1 and 0 le A le 90^(@)`, find :
`tan^(2)A-1/(cos^(2)A)`

Answer

To solve the problem, we need to find the value of \( \tan^2 A - \frac{1}{\cos^2 A} \) given that \( \sin 3A = 1 \) and \( 0 \leq A \leq 90^\circ \).

### Step-by-Step Solution:

1. **Understanding the given equation**:
We have \( \sin 3A = 1 \). The sine function equals 1 at \( 90^\circ \) and its odd multiples. Therefore, we can set:
\[ 3A = 90^\circ \]

2. **Solving for A**:
Dividing both sides by 3 gives:
\[ A = \frac{90^\circ}{3} = 30^\circ \]

3. **Finding \( \tan^2 A \)**:
Now, we need to calculate \( \tan^2 A \) when \( A = 30^\circ \):
\[ \tan 30^\circ = \frac{1}{\sqrt{3}} \quad \Rightarrow \quad \tan^2 30^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

4. **Finding \( \cos^2 A \)**:
Next, we calculate \( \cos^2 A \) when \( A = 30^\circ \):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \quad \Rightarrow \quad \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

5. **Calculating \( \frac{1}{\cos^2 A} \)**:
Now we find \( \frac{1}{\cos^2 A} \):
\[ \frac{1}{\cos^2 30^\circ} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

6. **Putting it all together**:
Now we substitute back into the expression:
\[ \tan^2 A - \frac{1}{\cos^2 A} = \frac{1}{3} - \frac{4}{3} \]
Simplifying this gives:
\[ = \frac{1 - 4}{3} = \frac{-3}{3} = -1 \]

### Final Answer:
Thus, the final answer is:
\[ \tan^2 A - \frac{1}{\cos^2 A} = -1 \]