Question

If $\mathrm{cos\theta }=-\frac{3}{5}\mathrm{and}\mathrm{\pi }<\mathrm{\theta }<\frac{3\mathrm{\pi }}{2}$ find the values of other five trigonometric functions and hence evaluate $\frac{\mathrm{cosec\theta }+\mathrm{cot\theta }}{\mathrm{sec\theta }-\mathrm{tan\theta }}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \cos \theta =-\frac{3}{5}\mathrm{and}\mathrm{\pi }<\mathrm{\theta }<\frac{3\mathrm{\pi }}{2} \\ \mathrm{Thus},\theta \mathrm{is}\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{quadrant}. \\ \mathrm{In}\mathrm{the}\mathrm{third}\mathrm{quadrant},\tan \theta \mathrm{and}\cot \theta \mathrm{are}\mathrm{positive} \\ \mathrm{And},\mathrm{all}\mathrm{the}\mathrm{other}\mathrm{four}\mathrm{T}-\mathrm{ratios}\mathrm{are}\mathrm{negative}. \\ \therefore \mathrm{sin\theta }=-\sqrt{1-{\cos }^2\mathrm{\theta }}=-\sqrt{1-{\left({\displaystyle \frac{-3}{5}}\right)}^2}=\frac{-4}{5} \\ \mathrm{tan\theta }=\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}=\frac{\frac{-4}{5}}{-\frac{3}{5}}=\frac{4}{3} \\ \mathrm{cot\theta }=\frac{1}{\mathrm{tan\theta }}=\frac{1}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}=\frac{3}{4} \\ \mathrm{sec\theta }=\frac{1}{\cos \mathrm{\theta }}=\frac{1}{{\displaystyle \raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}=\frac{-5}{3} \\ \mathrm{cosec\theta }=\frac{1}{\mathrm{sin\theta }}=\frac{1}{{\displaystyle \raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}=\frac{-5}{4} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\frac{\cos ec\mathrm{\theta }+\mathrm{cot\theta }}{sec\mathrm{\theta }-\mathrm{tan\theta }}=\frac{{\displaystyle \frac{-5}{4}}+{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{-5}{3}}-{\displaystyle \frac{4}{3}}} \\ =\frac{{\displaystyle \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{{\displaystyle \raisebox{1ex}{$-9$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \\ =\frac{{\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-3}=\frac{1}{6} \end{gathered}$$