Let α be the root of the equation 25 ^2 θ+5 θ-12=0 , where (π)/(2)...

Here $\alpha$ is the root of equation $25 \cos ^2 \theta+5 \cos \theta-12=0$$\begin{array}{ll}\Rightarrow \quad 25 \cos ^2 \alpha+5 \cos \alpha-12=0 \\ \Rightarrow \quad 25 \cos ^2 \alpha+20 \cos \alpha-15 \cos \alpha-12=0 \\ \Rightarrow \quad 5 \cos \alpha(5 \cos \alpha+4)-3(5 \cos \alpha+4)=0 \\ & (5 \cos \alpha-3)(5 \cos \alpha+4)=0 \\ & \cos \alpha=\frac{3}{5} \text{ or } \cos \alpha=\frac{-4}{5}\end{array}$Here, $\frac{\pi}{2}$ < \alpha < \pi$\cos \alpha=\frac{-4}{5} ( $\because$ In $2^{$\text{nd }$}$ quadrant, \cos \alpha$ value is negative) Now, $\sin \alpha=\sqrt{1-\cos$ ^2 \alpha}=\sqrt{1-$\frac{16}{25} }\begin{aligned} & \therefore \quad \sin \alpha=\frac{3}{5} \\ & \therefore \quad \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{3}{5} \times \frac{-5}{4}=\frac{-3}{4} \end{aligned}$$\therefore \quad$ Option (A) is correct.