Given that $\sin \theta = \frac{x}{y}$, we need to express $\sec \theta + \cot \theta$ in terms of $x$ and $y$.

Step 1: Use the Pythagorean Identity

Since $\sin \theta = \frac{x}{y}$, we know: $\cos^2 \theta + \sin^2 \theta = 1$ This implies: $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left( \frac{x}{y} \right)^2} = \sqrt{\frac{y^2 - x^2}{y^2}} = \frac{\sqrt{y^2 - x^2}}{y}$

Step 2: Express $\sec \theta$

The secant function is the reciprocal of cosine, so: $\sec \theta = \frac{1}{\cos \theta} = \frac{y}{\sqrt{y^2 - x^2}}$

Step 3: Express $\cot \theta$

The cotangent function is the ratio of cosine to sine, so: $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{\sqrt{y^2 - x^2}}{y}}{\frac{x}{y}} = \frac{\sqrt{y^2 - x^2}}{x}$

Step 4: Add $\sec \theta$ and $\cot \theta$

Now, combine the expressions for $\sec \theta$ and $\cot \theta$: $\sec \theta + \cot \theta = \frac{y}{\sqrt{y^2 - x^2}} + \frac{\sqrt{y^2 - x^2}}{x}$

This is the simplified expression for $\sec \theta + \cot \theta$ in terms of $x$ and $y$.

Would you like further details or have any questions?

Follow-up questions:

1. How does the Pythagorean identity relate to trigonometric functions?
2. Can you derive the relationship between secant and cosine?
3. Why is the cotangent function the ratio of cosine to sine?
4. Can this expression be simplified further for specific values of $x$ and $y$?
5. How do trigonometric identities help in solving complex expressions?

Tip:

Understanding the reciprocal identities (secant, cosecant, and cotangent) helps simplify many trigonometric problems quickly.