Question

If $cot\theta =-\frac{7}{24},\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{sin\theta }\mathrm{and}\mathrm{sec\theta },\mathrm{\theta }\mathrm{is}\mathrm{in}\mathrm{IV}\mathrm{quadrant}.$

Answer 1

$$\begin{gathered} \mathrm{Given}:cot\theta =-\frac{7}{24},\mathrm{i}.\mathrm{e}.,\theta \mathrm{lies}\mathrm{in}\mathrm{quadrant}\mathrm{IV}. \\ \therefore \tan \theta =\frac{1}{cot\theta }=-\frac{24}{7} \\ \mathrm{We}\mathrm{have}1+{\tan }^2\theta =se{c}^2\theta \\ \Rightarrow 1+{\left(-\frac{24}{7}\right)}^2=se{c}^2\theta \\ \Rightarrow se{c}^2\theta =\frac{625}{49} \\ \Rightarrow sec\theta =\pm \sqrt{\frac{625}{49}}=\frac{25}{7}(\because \theta \mathrm{lies}\mathrm{in}\mathrm{quadrant}\mathrm{IV}) \\ \mathrm{Now},\cos \theta =\frac{1}{sec\theta }=\frac{1}{{\displaystyle \frac{25}{7}}} \\ \Rightarrow \cos \theta =\frac{7}{25} \\ \mathrm{Also},{\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\sin }^2\theta +{\left(\frac{7}{25}\right)}^2=1 \\ \Rightarrow {\sin }^2\theta =1-{\left(\frac{7}{25}\right)}^2=\frac{576}{625} \\ \Rightarrow \sin \theta =\pm \sqrt{\frac{576}{625}}=-\frac{24}{25}(\because \theta \mathrm{lies}\mathrm{in}\mathrm{quadrant}\mathrm{IV}) \end{gathered}$$