Question

If $\sin \theta =\frac{5}{13},\mathrm{where}\theta$ is an acute angle , find the value of other trigonometric ratios using identities.

Answer 1

$$\begin{gathered} \mathrm{Given}:\sin \theta =\frac{5}{13} \\ \therefore \cos ec\theta =\frac{1}{\sin \theta }=\frac{1}{{\displaystyle \frac{5}{13}}} \\ \Rightarrow \cos ec\theta =\frac{13}{5} \\ \mathrm{We}\mathrm{have}{\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(\frac{5}{13}\right)}^2+{\cos }^2\theta =1 \\ \Rightarrow {\cos }^2\theta =1-{\left(\frac{5}{13}\right)}^2=\frac{144}{169} \\ \Rightarrow \cos \theta =\sqrt{\frac{144}{169}}=\frac{12}{13} \end{gathered}$$

$$\begin{gathered} \mathrm{Also},sec\theta =\frac{1}{\cos \theta }=\frac{1}{{\displaystyle \frac{12}{13}}} \\ \Rightarrow sec\theta =\frac{13}{12} \\ \mathrm{We}\mathrm{have}1+co{t}^2\theta =\cos e{c}^2\theta \\ \Rightarrow 1+co{t}^2\theta ={\left({\displaystyle \frac{13}{5}}\right)}^2 \\ \Rightarrow co{t}^2\theta =\frac{144}{25} \\ \Rightarrow cot\theta =\frac{12}{5} \\ \therefore \tan \theta =\frac{1}{cot\theta }=\frac{1}{{\displaystyle \frac{12}{5}}}=\frac{5}{12} \end{gathered}$$