If `sin21^(@)=(x)/(y)` , then `sec21^(@)-sin69^(@)` is equal to

Options

A. `(x^(2))/(ysqrt(y^(2)-x^(2)))`
B. `(y^(2))/(xsqrt(y^(2)-x^(2)))`
C. `(x^(2))/(ysqrt(x^(2)-y^(2)))`
D. `(y^(2))/(xsqrt(x^(2)-y^(2)))`

Answer

To solve the problem, we need to find the value of \( \sec 21^\circ - \sin 69^\circ \) given that \( \sin 21^\circ = \frac{x}{y} \).

### Step-by-Step Solution:

1. **Understanding the Relationship Between Angles**:
We know that \( \sin 69^\circ = \cos 21^\circ \) because \( 69^\circ + 21^\circ = 90^\circ \). Thus, we can rewrite the expression:
\[ \sec 21^\circ - \sin 69^\circ = \sec 21^\circ - \cos 21^\circ \]

2. **Expressing Secant and Cosine**:
The secant function is defined as:
\[ \sec 21^\circ = \frac{1}{\cos 21^\circ} \]
Therefore, we can rewrite our expression as:
\[ \sec 21^\circ - \sin 69^\circ = \frac{1}{\cos 21^\circ} - \cos 21^\circ \]

3. **Finding a Common Denominator**:
To combine the terms, we need a common denominator:
\[ = \frac{1 - \cos^2 21^\circ}{\cos 21^\circ} \]

4. **Using the Pythagorean Identity**:
We know from the Pythagorean identity that:
\[ 1 - \cos^2 \theta = \sin^2 \theta \]
Thus, we can substitute:
\[ = \frac{\sin^2 21^\circ}{\cos 21^\circ} \]

5. **Substituting the Value of \( \sin 21^\circ \)**:
Given that \( \sin 21^\circ = \frac{x}{y} \), we have:
\[ \sin^2 21^\circ = \left(\frac{x}{y}\right)^2 = \frac{x^2}{y^2} \]
Therefore, substituting this into our expression gives:
\[ = \frac{\frac{x^2}{y^2}}{\cos 21^\circ} \]

6. **Finding \( \cos 21^\circ \)**:
From our triangle, we can find \( \cos 21^\circ \) using the sides:
\[ \cos 21^\circ = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{y^2 - x^2}}{y} \]
Thus, we can rewrite the expression:
\[ = \frac{\frac{x^2}{y^2}}{\frac{\sqrt{y^2 - x^2}}{y}} = \frac{x^2}{y \sqrt{y^2 - x^2}} \]

7. **Final Result**:
Therefore, we conclude that:
\[ \sec 21^\circ - \sin 69^\circ = \frac{x^2}{y \sqrt{y^2 - x^2}} \]

### Final Answer:
The final answer is:
\[ \sec 21^\circ - \sin 69^\circ = \frac{x^2}{y \sqrt{y^2 - x^2}} \]