Given: \(\cos \alpha = \frac{5}{13}\) and \(0^\circ < \alpha < 90^\circ\) Prove that: \(\cot \alpha + \csc \alpha = 1.5\)

Q.Given: \(\cos \alpha = \frac{5}{13}\) and \(0^\circ < \alpha < 90^\circ\) Prove that: \(\cot \alpha + \csc \alpha = 1.5\)

\(\cot \alpha + \csc \alpha\) \[ = \frac{\cos \alpha}{\sin \alpha} + \frac{1}{\sin \alpha} = \frac{\cos \alpha + 1}{\sin \alpha} = \frac{\cos \alpha + 1}{\sqrt{1 - \cos^2 \alpha}} = \frac{\frac{5}{13} + 1}{\sqrt{1 - \left(\frac{5}{13}\right)^2}} = \frac{\frac{5 + 13}{13}}{\sqrt{\frac{169 - 25}{169}}} = \frac{\frac{18}{13}}{\sqrt{\frac{144}{169}}} = \frac{\frac{18}{13}}{\frac{12}{13}} = \frac{18}{13} \times \frac{13}{12} = \frac{3}{2} = 1.5 \] ( proved)