8.9 Given that \( 5 \cos \beta-3=0 \) and \( 0^{\circ}<\beta<90^{\circ} \). If \( \alpha+\beta=90^{\circ} \) and \( 0^{\circ}<\alpha<90^{\circ} \), calculate the value of \( \cot \alpha \).

To solve the equation \( 5 \cos \beta - 3 = 0 \), we first isolate \( \cos \beta \): \[ 5 \cos \beta = 3 \quad \Rightarrow \quad \cos \beta = \frac{3}{5} \] Next, since \( \alpha + \beta = 90^{\circ} \), we know that \( \alpha \) and \( \beta \) are complementary angles. Therefore, we can express \( \sin \alpha \) in terms of \( \cos \beta \): \[ \sin \alpha = \cos \beta = \frac{3}{5} \] To find \( \cot \alpha \), we use the identity: \[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \] First, we need to find \( \cos \alpha \). Knowing that \( \sin^2 \alpha + \cos^2 \alpha = 1\): \[ \sin^2 \alpha = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \] We can find \( \cos^2 \alpha \): \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \] Thus, we have: \[ \cos \alpha = \sqrt{\frac{16}{25}} = \frac{4}{5} \] Now, substituting \( \sin \alpha \) and \( \cos \alpha \) into \( \cot \alpha \): \[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \] Therefore, the value of \( \cot \alpha \) is: \[ \boxed{\frac{4}{3}} \]