Let P(x,y) denote the point where the terminal side of an angle θ meets the unit circle. If P is in Quadrant II and y=5/8, find tan(θ)+sec(θ).

The point \( P(x, y) \) lies on the unit circle, meaning \( P(x, y) \) satisfies the equation of the unit circle:

\[
x^2 + y^2 = 1
\]

Given that \( y = \frac{5}{8} \), substitute this into the equation:

\[
x^2 + \left( \frac{5}{8} \right)^2 = 1
\]

Simplifying:

\[
x^2 + \frac{25}{64} = 1
\]

Subtract \( \frac{25}{64} \) from both sides:

\[
x^2 = 1 - \frac{25}{64} = \frac{64}{64} - \frac{25}{64} = \frac{39}{64}
\]

Thus:

\[
x^2 = \frac{39}{64}
\]

Taking the square root of both sides:

\[
x = \pm \frac{\sqrt{39}}{8}
\]

Since the point \( P(x, y) \) is in Quadrant II, the x-coordinate must be negative. Therefore, we take the negative root:

\[
x = -\frac{\sqrt{39}}{8}
\]

Now, we can compute \( \tan(\theta) \) and \( \sec(\theta) \):

1. **Tangent**: 
   \[
   \tan(\theta) = \frac{y}{x} = \frac{\frac{5}{8}}{-\frac{\sqrt{39}}{8}} = -\frac{5}{\sqrt{39}}
   \]

2. **Secant**: 
   \[
   \sec(\theta) = \frac{1}{x} = \frac{1}{-\frac{\sqrt{39}}{8}} = -\frac{8}{\sqrt{39}}
   \]

Finally, compute \( \tan(\theta) + \sec(\theta) \):

\[
\tan(\theta) + \sec(\theta) = -\frac{5}{\sqrt{39}} + \left( -\frac{8}{\sqrt{39}} \right) = -\frac{5 + 8}{\sqrt{39}} = -\frac{13}{\sqrt{39}}
\]

Rationalizing the denominator:

\[
-\frac{13}{\sqrt{39}} \times \frac{\sqrt{39}}{\sqrt{39}} = -\frac{13\sqrt{39}}{39}
\]

Thus, the final result is:

\[
\tan(\theta) + \sec(\theta) = -\frac{13\sqrt{39}}{39}
\]

Would you like further details or clarifications?

Here are five related questions:
1. What is the general formula for secant in terms of \( x \) and \( y \) on the unit circle?
2. Why is the x-coordinate negative in Quadrant II?
3. How do you rationalize a denominator in an expression like \( \frac{13}{\sqrt{39}} \)?
4. How does the value of \( \tan(\theta) \) vary between different quadrants?
5. Can you derive sine and cosine from the coordinates \( P(x, y) \) on the unit circle?

**Tip**: Always use the quadrant information to determine the sign of trigonometric functions when solving problems involving the unit circle.

Solve for tan(θ) + sec(θ) when the point P(x, y) is in Quadrant II on the unit circle with y = 5/8. This problem involves understanding trigonometric functions, the unit circle, and rationalizing expressions. Suitable for students in Grades 10-12.

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