Key Concepts and Formulas

• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Trigonometric Identity: $\sec A = \frac{1}{\cos A}$.
• Pythagorean Identity: $\sec^2 A = 1 + \tan^2 A$.
• Trigonometric Signs in Quadrants: In the second quadrant ($\frac{\pi}{2} < A < \pi$), cosine (and secant) are negative, and tangent is also negative.

Step-by-Step Solution

Step 1: Determine the Value of $\cos A$

We are given the equation $5 \cos A + 3 = 0$. Why this step? This allows us to find the value of $\cos A$, which is crucial for finding other trigonometric functions.

Solving for $\cos A$: $5 \cos A = -3$ $\cos A = -\frac{3}{5}$

Step 2: Determine the Value of $\sec A$

We use the reciprocal identity $\sec A = \frac{1}{\cos A}$. Why this step? The options include $\sec A$, so we need to calculate its value.

Substituting the value of $\cos A$: $\sec A = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}$

Step 3: Solve the Quadratic Equation

The quadratic equation is $9x^2 + 27x + 20 = 0$. Why this step? We need to find the roots of this equation to compare them with the trigonometric values.

Using the quadratic formula with $a=9$, $b=27$, and $c=20$: $x = \frac{-27 \pm \sqrt{27^2 - 4(9)(20)}}{2(9)}$ $x = \frac{-27 \pm \sqrt{729 - 720}}{18}$ $x = \frac{-27 \pm \sqrt{9}}{18}$ $x = \frac{-27 \pm 3}{18}$

The two roots are: $x_1 = \frac{-27 + 3}{18} = \frac{-24}{18} = -\frac{4}{3}$ $x_2 = \frac{-27 - 3}{18} = \frac{-30}{18} = -\frac{5}{3}$

So, the roots are $-\frac{4}{3}$ and $-\frac{5}{3}$.

Step 4: Determine the Value of $\tan A$

We use the Pythagorean identity $\sec^2 A = 1 + \tan^2 A$. Why this step? The options include $\tan A$, so we need to find its value.

Rearranging the identity: $\tan^2 A = \sec^2 A - 1$ Substituting the value of $\sec A = -\frac{5}{3}$: $\tan^2 A = \left(-\frac{5}{3}\right)^2 - 1 = \frac{25}{9} - 1 = \frac{16}{9}$ $\tan A = \pm \sqrt{\frac{16}{9}} = \pm \frac{4}{3}$

Since $A$ is an angle of a triangle and $\cos A < 0$, $A$ must be in the second quadrant. In the second quadrant, $\tan A$ is negative. Therefore: $\tan A = -\frac{4}{3}$

Step 5: Match the Roots with the Trigonometric Values

We found the roots of the quadratic equation to be $-\frac{4}{3}$ and $-\frac{5}{3}$. We found $\sec A = -\frac{5}{3}$ and $\tan A = -\frac{4}{3}$. Why this step? This is the final step where we directly compare our calculated roots with our calculated trigonometric function values to identify which trigonometric functions correspond to the roots.

Therefore, the roots of the quadratic equation are $\sec A$ and $\tan A$.

Common Mistakes & Tips

• Sign of $\tan A$: Remember to consider the quadrant of angle $A$ when determining the sign of $\tan A$. Since $\cos A$ is negative, $A$ is in the second quadrant where $\tan A$ is negative.
• Using the Correct Identity: Make sure to use the correct trigonometric identities. A small mistake here can lead to an incorrect answer.
• Arithmetic Errors: Be careful with arithmetic, especially when using the quadratic formula.

Summary

We first determined the value of $\cos A$ from the given equation and then found $\sec A$. Next, we solved the quadratic equation to find its roots. Finally, using the Pythagorean identity and considering the quadrant of $A$, we found $\tan A$. Comparing these values, we concluded that the roots of the quadratic equation are $\sec A$ and $\tan A$.

The final answer is \boxed{secA, tanA}, which corresponds to option (A).