Question

Find the values of the other five trigonometric functions in each of the following:

(i) $\cot x=\frac{12}{5},$ x in quadrant III

(ii) $\cos x=-\frac{1}{2},$ x in quadrant II

(iii) $\tan x=\frac{3}{4},$ x in quadrant III

(iv) $\sin x=\frac{3}{5},$ x in quadrant I

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}: \\ \cot x=\frac{12}{5}\mathrm{and}x\text{are}\mathrm{in}\text{the third quadrant.} \\ \mathrm{In}\text{the third quadrant,}\tan x\mathrm{and}\cot x\mathrm{are}\mathrm{positive} \\ \text{A}\mathrm{nd},\sin x,\cos x,\sec x\mathrm{and}\mathrm{cosec}x\mathrm{are}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore \tan x=\frac{1}{\cot x}=\frac{1}{{\displaystyle \frac{12}{5}}}=\frac{5}{12} \\ \mathrm{cosec}x=-\sqrt{1+co{t}^{2}x}=-\sqrt{1+{\left({\displaystyle \frac{12}{5}}\right)}^2}=-\frac{13}{5} \\ \sin x=\frac{1}{\mathrm{cosec}x}=\frac{1}{-\frac{13}{5}}=-\frac{5}{13} \\ \cos x=-\sqrt{1-{\sin }^{2}x}=-\sqrt{1-{\left({\displaystyle \frac{-5}{13}}\right)}^2}=\frac{-12}{13} \\ \mathrm{And},\sec x=\frac{1}{\cos x}=\frac{1}{{\displaystyle \frac{-12}{13}}}=\frac{-13}{12} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}: \\ \cos x=-\frac{1}{2}\mathrm{and}x\text{are in the second quadrant.} \\ \mathrm{In}\text{the second quadrant,}\sin x,\mathrm{and}\mathrm{cosec}x\mathrm{are}\mathrm{positive}. \\ \mathrm{And},\tan x,\cot x,\cos x\mathrm{and}\sec x\mathrm{are}\mathrm{negative}. \end{gathered}$$

$$\begin{gathered} \therefore \sin x=\sqrt{1-{\cos }^{2}x}=\sqrt{1-{\left({\displaystyle \frac{-1}{2}}\right)}^2}=\frac{\sqrt{3}}{2} \\ \tan x=\frac{\sin x}{\cos x}=\frac{{\displaystyle \raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=-\sqrt{3} \end{gathered}$$
$$\begin{gathered} \mathrm{c}\mathrm{o}\mathrm{t}x=\frac{1}{\tan x}=\frac{1}{-\sqrt{3}}=\frac{-1}{\sqrt{3}} \\ \sec x=\frac{1}{\cos x}=\frac{1}{{\displaystyle \frac{-1}{2}}}=-2 \\ \mathrm{cosec}x=\frac{1}{\sin x}=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}: \\ \tan x=\frac{3}{4}\mathrm{and}x\text{are in the third quadrant.} \\ \mathrm{In}\text{the third quadrant,}\tan x,\mathrm{and}\cot x\mathrm{are}\mathrm{positive}. \\ \mathrm{And},\sin x,\cos x,\sec x\mathrm{and}\mathrm{cosec}x\mathrm{are}\mathrm{negative}. \end{gathered}$$
$$\begin{gathered} \therefore \mathrm{c}\mathrm{o}\mathrm{t}x=\frac{1}{\tan x}=\frac{1}{{\displaystyle \frac{3}{4}}}=\frac{4}{3} \\ \sec x=-\sqrt{1+{\tan }^{2}x}=-\sqrt{1+{\left({\displaystyle \frac{3}{4}}\right)}^2}=-\frac{5}{4} \\ \cos x=\frac{1}{\sec x}=\frac{1}{-\frac{5}{4}}=-\frac{4}{5} \\ \sin x=-\sqrt{1-{\cos }^{2}x}=-\sqrt{1-{\left({\displaystyle \frac{-4}{5}}\right)}^2}=\frac{-3}{5} \\ \mathrm{cosec}x=\frac{1}{\sin x}=\frac{1}{-\frac{3}{5}}=-\frac{5}{3} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}: \\ \sin x=\frac{3}{5}\mathrm{and}x\text{are in the first quadrant.} \\ \mathrm{In}\text{the first quadrant,}\mathrm{all}\mathrm{six}\mathrm{T}-\mathrm{ratios}\mathrm{are}\mathrm{positive}. \end{gathered}$$
$$\begin{gathered} \therefore \cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-{\left({\displaystyle \frac{{\displaystyle 3}}{5}}\right)}^2}=\frac{4}{5} \\ \tan x=\frac{\sin x}{\cos x}=\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}=\frac{3}{4} \\ \cot x=\frac{1}{\tan x}=\frac{1}{\frac{3}{4}}=\frac{4}{3} \\ \sec x=\frac{1}{\cos x}=\frac{1}{{\displaystyle \frac{4}{5}}}=\frac{5}{4} \\ \mathrm{cosec}x=\frac{1}{\sin x}=\frac{1}{{\displaystyle \frac{3}{5}}}=\frac{5}{3} \end{gathered}$$