To solve for [tex]\(\tan^2 A + 1\)[/tex] given the equation [tex]\(5 \cos A + 3 = 0\)[/tex] and the condition that [tex]\(180^\circ < A < 360^\circ\)[/tex], we can follow these steps:
1. Solve for [tex]\(\cos A\)[/tex]:
Rearrange the given equation:
[tex]\[
5 \cos A + 3 = 0
\][/tex]
[tex]\[
\cos A = -\frac{3}{5}
\][/tex]
2. Determine the quadrant:
Since [tex]\(180^\circ < A < 360^\circ\)[/tex], angle [tex]\(A\)[/tex] is either in the third quadrant (where both sine and tangent are positive) or the fourth quadrant (where sine is negative and tangent is negative).
3. Calculate [tex]\(\sin A\)[/tex] using the Pythagorean identity:
[tex]\[
\sin^2 A + \cos^2 A = 1
\][/tex]
Substitute [tex]\(\cos A = -\frac{3}{5}\)[/tex]:
[tex]\[
\sin^2 A + \left( -\frac{3}{5} \right)^2 = 1
\][/tex]
[tex]\[
\sin^2 A + \frac{9}{25} = 1
\][/tex]
[tex]\[
\sin^2 A = 1 - \frac{9}{25}
\][/tex]
[tex]\[
\sin^2 A = \frac{25}{25} - \frac{9}{25}
\][/tex]
[tex]\[
\sin^2 A = \frac{16}{25}
\][/tex]
Therefore, [tex]\(\sin A\)[/tex] can be:
[tex]\[
\sin A = \pm \frac{4}{5}
\][/tex]
In the third quadrant, [tex]\(\sin A\)[/tex] is positive:
[tex]\[
\sin A = \frac{4}{5}
\][/tex]
In the fourth quadrant, [tex]\(\sin A\)[/tex] is negative:
[tex]\[
\sin A = -\frac{4}{5}
\][/tex]
4. Calculate [tex]\(\tan A\)[/tex]:
[tex]\[
\tan A = \frac{\sin A}{\cos A}
\][/tex]
- In the third quadrant:
[tex]\[
\tan A = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}
\][/tex]
- In the fourth quadrant:
[tex]\[
\tan A = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{4}{3}
\][/tex]
5. Calculate [tex]\(\tan^2 A + 1\)[/tex]:
[tex]\[
\tan^2 A = \left( \frac{4}{3} \right)^2 = \frac{16}{9}
\][/tex]
[tex]\[
\tan^2 A + 1 = \frac{16}{9} + 1
\][/tex]
Convert 1 to a fraction with a denominator of 9:
[tex]\[
\tan^2 A + 1 = \frac{16}{9} + \frac{9}{9} = \frac{25}{9}
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\frac{25}{9}}
\][/tex]
