Question

If sin $\mathrm{\theta }=\frac{12}{13}$ and θ lies in the second quadrant, find the value of sec θ + tan θ.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ \sin \theta =\frac{12}{13}\mathrm{and}\theta \mathrm{lie}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant}. \\ \mathrm{In}\mathrm{the}\mathrm{second}\mathrm{quadrant},\sin \theta \mathrm{and}\mathrm{cosec}\theta \mathrm{are}\mathrm{positive}\mathrm{and}\mathrm{all}\mathrm{the}\mathrm{other}\mathrm{four}\mathrm{T}-\mathrm{ratios}\mathrm{are}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore \cos \theta =-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-{\left(\frac{12}{13}\right)}^2} \\ =\frac{-5}{13} \\ \tan \theta =\frac{\sin \theta }{\cos \theta } \\ =\frac{{\displaystyle \frac{12}{13}}}{{\displaystyle \frac{-5}{13}}} \\ =\frac{-12}{5} \\ \text{And, sec}\theta =\frac{1}{\cos \theta } \\ =\frac{1}{{\displaystyle \frac{-5}{13}}} \\ =\frac{-13}{5} \\ \therefore \text{sec}\theta +\tan \theta =\frac{-13}{5}+\frac{-12}{5} \\ =-5 \end{gathered}$$