Question

$3\sin \alpha -4\cos \propto =0,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}\tan \alpha ,sec\alpha ,\cos ec\alpha ,\mathrm{where}\mathrm{\alpha }\mathrm{is}\mathrm{an}\mathrm{acute}\mathrm{angle}.$

Answer 1

$$\begin{gathered} \mathrm{Given}:3\sin \alpha -4\cos \alpha =0 \\ \Rightarrow 3\sin \alpha =4\cos \alpha \\ \Rightarrow \frac{\sin \alpha }{\cos \alpha }=\frac{4}{3} \\ \Rightarrow \tan \alpha =\frac{4}{3} \\ \mathrm{We}\mathrm{have}1+{\tan }^2\alpha =se{c}^2\alpha \\ \Rightarrow 1+{\left({\displaystyle \frac{4}{3}}\right)}^2=se{c}^2\alpha \\ \Rightarrow se{c}^2\alpha =\frac{25}{9} \\ \Rightarrow sec\alpha =\frac{5}{3}(\because \mathrm{\alpha }\mathrm{is}\mathrm{an}\mathrm{acute}\mathrm{angle}) \\ \mathrm{Also},1+co{t}^2\alpha =\cos e{c}^2\alpha \\ \Rightarrow 1+{\left(\frac{1}{\tan \alpha }\right)}^2=\cos e{c}^2\alpha \\ \Rightarrow 1+{\left(\frac{1}{{\displaystyle \frac{4}{3}}}\right)}^2=\cos e{c}^2\alpha \\ \Rightarrow 1+{\left(\frac{3}{4}\right)}^2=\cos e{c}^2\alpha \\ \Rightarrow \cos e{c}^2\alpha =\frac{25}{16} \\ \Rightarrow \cos ec\alpha =\frac{5}{4} \end{gathered}$$