If `sin17^(@)=(a)/(b)` then value of `sec17^(@)-sin73^(@)` is

Options

A. `(a)/(bsqrt(a^(2)+b^(2)))`
B. `(b^(2))/(asqrt(b^(2)-a^(2)))`
C. `(a^(2))/(bsqrt(b^(2)-a^(2)))`
D. `0`

Answer

To solve the problem, we need to find the value of \( \sec 17^\circ - \sin 73^\circ \) given that \( \sin 17^\circ = \frac{a}{b} \).

### Step-by-Step Solution:

1. **Express \( \sec 17^\circ \)**:
\[ \sec 17^\circ = \frac{1}{\cos 17^\circ} \]

2. **Use the co-function identity**:
We know that \( \sin(90^\circ - x) = \cos x \). Thus,
\[ \sin 73^\circ = \cos 17^\circ \]

3. **Substituting \( \sin 73^\circ \)**:
Now we can rewrite the expression:
\[ \sec 17^\circ - \sin 73^\circ = \frac{1}{\cos 17^\circ} - \cos 17^\circ \]

4. **Finding a common denominator**:
The common denominator for the two terms is \( \cos 17^\circ \):
\[ \sec 17^\circ - \sin 73^\circ = \frac{1 - \cos^2 17^\circ}{\cos 17^\circ} \]

5. **Using the Pythagorean identity**:
We know that \( 1 - \cos^2 x = \sin^2 x \). Therefore:
\[ 1 - \cos^2 17^\circ = \sin^2 17^\circ \]

6. **Substituting back**:
Now we substitute this back into our expression:
\[ \sec 17^\circ - \sin 73^\circ = \frac{\sin^2 17^\circ}{\cos 17^\circ} \]

7. **Substituting \( \sin 17^\circ \)**:
Given that \( \sin 17^\circ = \frac{a}{b} \), we have:
\[ \sin^2 17^\circ = \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2} \]

8. **Finding \( \cos 17^\circ \)**:
We can find \( \cos 17^\circ \) using the identity \( \cos^2 17^\circ = 1 - \sin^2 17^\circ \):
\[ \cos^2 17^\circ = 1 - \frac{a^2}{b^2} = \frac{b^2 - a^2}{b^2} \]
Thus,
\[ \cos 17^\circ = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b} \]

9. **Final substitution**:
Now substituting \( \cos 17^\circ \) into our expression:
\[ \sec 17^\circ - \sin 73^\circ = \frac{\frac{a^2}{b^2}}{\frac{\sqrt{b^2 - a^2}}{b}} = \frac{a^2}{b \sqrt{b^2 - a^2}} \]

### Final Answer:
Thus, the value of \( \sec 17^\circ - \sin 73^\circ \) is:
\[ \frac{a^2}{b \sqrt{b^2 - a^2}} \]