Question

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

Answer 1

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