1. Given condition on the angle

We have $5\cos A + 3 = 0$ So, $\cos A = -\frac{3}{5}$

Since $A$ is an angle of a triangle, $0<A<\pi$. Also $\cos A<0$, so $A$ is obtuse and hence $\sin A>0$.

Therefore, $\sin A = \sqrt{1-\cos^2 A} = \sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Now compute the required trigonometric values: $\sec A = \frac{1}{\cos A} = -\frac{5}{3}$ $\tan A = \frac{\sin A}{\cos A} = \frac{4/5}{-3/5} = -\frac{4}{3}$ $\cot A = \frac{1}{\tan A} = -\frac{3}{4}$

2. Solve the quadratic equation

Given: $9x^2+27x+20=0$

Factorize: $9x^2+27x+20 = 9x^2+15x+12x+20$ $=3x(3x+5)+4(3x+5)$ $=(3x+5)(3x+4)$

So the roots are $x=-\frac{5}{3},\; -\frac{4}{3}$

3. Match with trigonometric values

We found: $\sec A=-\frac{5}{3}, \qquad \tan A=-\frac{4}{3}$

These are exactly the roots of the quadratic equation.

So the roots are: $\sec A,\; \tan A$

4. Check options

• A: $\sec A, \cot A = -\frac{5}{3}, -\frac{3}{4}$ ❌
• B: $\sin A, \sec A = \frac{4}{5}, -\frac{5}{3}$ ❌
• C: $\sec A, \tan A = -\frac{5}{3}, -\frac{4}{3}$ ✅
• D: $\tan A, \cos A = -\frac{4}{3}, -\frac{3}{5}$ ❌

Hence, the correct option is C.