Question

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

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Solution

$\mathrm{We}\mathrm{have}:\phantom{\rule{0ex}{0ex}}\mathrm{sin}\theta =\frac{12}{13}\mathrm{and}\theta \mathrm{lie}\mathrm{in}\mathrm{the}\mathrm{second}\mathrm{quadrant}.\phantom{\rule{0ex}{0ex}}\mathrm{In}\mathrm{the}\mathrm{second}\mathrm{quadrant},\mathrm{sin}\theta \mathrm{and}\mathrm{cosec}\theta \mathrm{are}\mathrm{positive}\mathrm{and}\mathrm{all}\mathrm{the}\mathrm{other}\mathrm{four}\mathrm{T}-\mathrm{ratios}\mathrm{are}\mathrm{negative}.$ $\therefore \mathrm{cos}\theta =-\sqrt{1-{\mathrm{sin}}^{2}\theta }\phantom{\rule{0ex}{0ex}}=-\sqrt{1-{\left(\frac{12}{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{-5}{13}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{\frac{12}{13}}{\frac{-5}{13}}\phantom{\rule{0ex}{0ex}}=\frac{-12}{5}\phantom{\rule{0ex}{0ex}}\text{And, sec}\theta =\frac{1}{\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{-5}{13}}\phantom{\rule{0ex}{0ex}}=\frac{-13}{5}\phantom{\rule{0ex}{0ex}}\therefore \text{sec}\theta +\mathrm{tan}\theta =\frac{-13}{5}+\frac{-12}{5}\phantom{\rule{0ex}{0ex}}=-5$

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