Question

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

Answer 1

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