$\begin{align} & \text{if }\cos \theta =\tfrac{4}{5}\text{ (i) }\sin \theta \text{ (ii) }\operatorname{cosec}\theta \text{ (iii) }\theta \\ & \text{If }\sin \theta =-\tfrac{12}{13}\text{ (i) cos}\theta \text{ (ii) }\tan \theta \\\end{align}$
$\begin{align} & \cos \theta =\frac{4}{5} \\ & \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }=\sqrt{1-{{(\tfrac{4}{5})}^{2}}}=\sqrt{1-\tfrac{16}{25}} \\ & \sin \theta =\sqrt{\frac{9}{25}}=\frac{3}{5} \\ & \operatorname{cosec}\theta =\frac{1}{\sin \theta }=\frac{1}{\tfrac{3}{5}}=\frac{5}{3} \\ & \theta ={{\sin }^{-1}}(\tfrac{3}{5})={{36}^{\circ }}\pm .52' \\ & \sin \theta =-\frac{12}{13} \\ & \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{(-\tfrac{12}{13})}^{2}}} \\ & \cos \theta =\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\pm \sqrt{\frac{5}{13}} \\ & \tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{{-12}/{13}\;}{\pm {5}/{13}\;}=\pm \frac{12}{5} \\\end{align}$