Question

If sin $\mathrm{\theta }=\frac{3}{5},\mathrm{tan\varphi }=\frac{1}{2}\mathrm{and}\frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\mathrm{\pi }<\mathrm{\varphi }<\frac{3\mathrm{\pi }}{2},$ find the value of 8 tan $\mathrm{\theta }-\sqrt{5}\sec \mathrm{\varphi }$.

Answer 1

We have:
$$\begin{gathered} \mathrm{sin\theta }=\frac{3}{5},\mathrm{tan\varphi }=\frac{1}{2}\mathrm{and}\frac{\mathrm{\pi }}{2}<\mathrm{\theta }<\mathrm{\pi }<\mathrm{\varphi }<\frac{3\mathrm{\pi }}{2}, \\ \mathrm{Thus},\mathrm{\theta }\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant}\mathrm{and}\mathrm{\varphi }\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant}. \\ \mathrm{In}\mathrm{the}\mathrm{second}\mathrm{quadrant},\mathrm{cos\theta }\mathrm{and}\mathrm{tan\theta }\mathrm{are}\mathrm{negative}. \\ \mathrm{In}\mathrm{the}\mathrm{third}\mathrm{quadrant},\mathrm{sec\varphi }\mathrm{is}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore \cos \theta =-\sqrt{1-{\sin }^2\theta }=-\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\frac{-4}{5} \\ \tan \theta =\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{{\displaystyle \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}=\frac{-3}{4} \\ \mathrm{And},\mathrm{sec\varphi }=-\sqrt{1+{\tan }^2\mathrm{\varphi }}=-\sqrt{1+{\left(\frac{1}{2}\right)}^2}=\frac{-\sqrt{5}}{2} \\ \therefore 8\tan \mathrm{\theta }-\sqrt{5}\mathrm{sec\varphi }=8\times \frac{-3}{4}-\sqrt{5}\times \frac{-\sqrt{5}}{2}=-6+\frac{5}{2}=-\frac{7}{2} \end{gathered}$$