Question 1: Find the values of the other five trigonometric functions in each of the following: $\displaystyle \text{i) } \cot x = \frac{12}{5} , x \text{ in Quadrant III } \hspace{1.0cm} \text{ii) } \cos x = \frac{-1}{2} , x \text{ in Quadrant II }$ $\displaystyle \text{iii) } \tan x = \frac{3}{4} , x \text{ in Quadrant III } \hspace{1.0cm} \text{iv) } \sin x = \frac{3}{5} , x \text{ in Quadrant I }$ $\displaystyle \text{i) } \cot x = \frac{12}{5} , x \text{ in Quadrant III }$ $\displaystyle \text{We know } \mathrm{cosec}^2 - \cot^2 x = 1 \Rightarrow \mathrm{cosec} x = \pm \sqrt{1 + \cot^2 x}$ $\displaystyle \text{In Quadrant III, } \mathrm{cosec} x \text{ is negative }$ $\displaystyle \therefore \mathrm{cosec} x = - \sqrt{1 + (\frac{12}{5})^2} = - \sqrt{\frac{169}{25}} = - \frac{13}{5}$ $\displaystyle \text{We know, } \tan x = \frac{1}{\cot x} = \frac{5}{12}$ $\displaystyle \text{Also } \sin x = \frac{1}{\mathrm{cosec} x} = - \frac{5}{13}$ $\displaystyle \text{Now } \sin^2 x + \cos^2 x = 1 \Rightarrow \cos x = \pm \sqrt{1 - \sin^2 x}$ $\displaystyle \text{In Q III, } \cos x$ is negative $\displaystyle \therefore \cos x = - \sqrt{1 - (\frac{-5}{13})^2} = - \sqrt{\frac{144}{169}} = - \frac{12}{13}$ $\displaystyle \therefore \sec x = \frac{1}{\cos x} = - \frac{13}{12}$ $\displaystyle \text{ii) } \cos x = \frac{-1}{2} , x \text{ in Quadrant II }$ $\displaystyle \text{We know } \sin^2 x + \cos^2 x = 1 \Rightarrow \cos x = \pm \sqrt{1 - \sin^2 x}$ $\displaystyle \text{In Quadrant II, } \sin x \text{ is positive } \text{and } \tan x \text{ is negative }$ $\displaystyle \therefore \sin x = \sqrt{1 - (\frac{1}{2})^2 } = \frac{\sqrt{3}}{2}$ $\displaystyle \therefore \mathrm{cosec} x = \frac{1}{\sin x} = \frac{2}{\sqrt{3}}$ $\displaystyle \tan x = \frac{\sin x }{\cos x} = \frac{\sqrt{3}}{2} (-2) = - \sqrt{3}$ $\displaystyle \therefore \cot x = \frac{1}{\tan x} = - \frac{1}{\sqrt{3}}$ $\displaystyle \sec x = \frac{1}{\cos x} = \frac{1}{-\frac{1}{2}} = -2$ $\displaystyle \text{iii) } \tan x = \frac{3}{4} , x \text{ in Quadrant III }$ $\displaystyle \therefore \cot x = \frac{1}{\tan x} = \frac{4}{3}$ $\displaystyle \text{In Quadrant III, } \mathrm{cosec} x \text{ is negative }$ $\displaystyle \therefore \mathrm{cosec} = - \sqrt{1 + \cot^2 x} = - \sqrt{1 + (\frac{4}{3})^2} = - \frac{5}{3}$ $\displaystyle \therefore \sin x = \frac{1}{\mathrm{cosec} x} = - \frac{3}{5}$ $\displaystyle \text{Also } \tan x = \frac{\sin x}{\cos x} \Rightarrow \cos x = - \frac{3}{5} . \frac{4}{3} = - \frac{4}{5}$ $\displaystyle \therefore \sec x = \frac{1}{\cos x} = - \frac{5}{4}$ $\displaystyle \text{iv) } \sin x = \frac{3}{5} , x \text{ in Quadrant I }$ $\displaystyle \text{In Quadrant I, } \cos x \text{ is positive }$ $\displaystyle \therefore \cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ $\displaystyle \therefore \tan x = \frac{\sin x}{\cos x} = \frac{3}{5} \times \frac{5}{4} = \frac{3}{4}$ $\displaystyle \therefore \cot x = \frac{1}{\tan x} = \frac{4}{3}$ $\displaystyle \text{and } \mathrm{cosec} = \frac{1}{\sin x} = \frac{5}{3}$ $\displaystyle \\$ $\displaystyle \text{Question 2: If } \sin x = \frac{12}{13} \text{and } x \text{lies in Quadrant II, find the value of } \\ \\ \sec x + \tan x .$ $\displaystyle \sin x = \frac{12}{13} , x$ in Quadrant II $\displaystyle \cos x$ in Quadrant II is negative $\displaystyle \therefore \cos x = - \sqrt{1 - \sin^2 x} = - \sqrt{1 - \frac{144}{169} } = - \frac{5}{13}$ $\displaystyle \therefore \sec x = \frac{1}{\cos x} = - \frac{13}{5}$ $\displaystyle \tan x= \frac{\sin x}{\cos x} = \frac{12}{13} \times \frac{-13}{5} = - \frac{12}{5}$ $\displaystyle \therefore \sec x + \tan x = - \frac{13}{5} + \frac{-12}{5} = - \frac{25}{5} = -5$ $\displaystyle \\$ $\displaystyle \text{Question 3: If } \sin x = \frac{3}{5} \text{and } \tan y = \frac{1}{2} \text{and } \frac{\pi}{2} < x < \pi < y < \frac{3\pi}{2}, \\ \\ \text{ find the value of } 8 \tan x - \sqrt{5} \sec y.$ $\displaystyle \text{Given } \sin x = \frac{3}{5} \text{and } \tan y = \frac{1}{2} \text{and } \frac{\pi}{2} < x < \pi < y < \frac{3\pi}{2}$ $\displaystyle \Rightarrow x$ in Quadrant II $\displaystyle \text{and } y$ in Quadrant III $\displaystyle \cos x$ in Quadrant II is negative $\displaystyle \therefore \cos x = \sqrt{1 - \sin^2 x} = - \sqrt{1 - \frac{9}{25}} = \frac{-4}{5}$ $\displaystyle \therefore \tan x= \frac{\sin x}{\cos x} = \frac{3}{5} \times \frac{-5}{4} = - \frac{3}{4}$ $\displaystyle \tan y = \frac{1}{2}$ $\displaystyle \sec^2 y = 1 + \tan^2 y = 1 + \Big( \frac{1}{2} \Big)^2 = \frac{5}{4}$ $\displaystyle \sec y$ in Quadrant III is negative $\displaystyle \therefore \sec y = - \sqrt{\frac{5}{4}} = - \frac{\sqrt{5}}{2}$ $\displaystyle \therefore 8 \tan x - \sqrt{5} \sec y = 8 \times (- \frac{3}{4} ) - \sqrt{5} (- \frac{\sqrt{5}}{2} ) = -6 + \frac{5}{2} = - \frac{7}{2}$ $\displaystyle \\$ $\displaystyle \text{Question 4: If } \sin x + \cos y = 0$, $\displaystyle \text{and } x \text{ lies in the fourth quadrant, find } \sin x \text{and } \cos x .$ $\displaystyle \sin x + \cos x = 0, x$ in Quadrant IV $\displaystyle \Rightarrow \sin x = - \cos x$ $\displaystyle \Rightarrow \tan x = - 1$ $\displaystyle \therefore \sec^2 x = 1 + \tan^2 x$ $\displaystyle \text{In Quadrant IV, } \sec x$ is positive $\displaystyle \therefore \sec x = \sqrt{1 + (-1)^2} = \sqrt{2}$ $\displaystyle \therefore \cos x = \frac{1}{\sec x} = \frac{1}{\sqrt{2}}$ $\displaystyle \text{Since } \tan x = \frac{\sin x }{\cos x}$ $\displaystyle \Rightarrow \sin x = -1 \times \cos x = -1 \times \frac{1}{\sqrt{2}} = - \frac{1}{\sqrt{2}}$ $\displaystyle \\$ $\displaystyle \text{Question 5: If } \cos x = - \frac{3}{5} \text{and } \pi < x < \frac{3\pi}{2}, \\ \\ \text{find the value of the five trigonometric functions and hence evaluate } \frac{\mathrm{cosec} x + \cot x}{\sec x - \tan x}.$ $\displaystyle \text{Given } \cos x = - \frac{3}{5} \text{and } \pi < x < \frac{3\pi}{2}$ $\displaystyle \Rightarrow x$ in Quadrant III $\displaystyle \sec x = \frac{1}{\cos x} = - \frac{5}{3}$ $\displaystyle \text{In Quadrant III, } \sin x$ is negative $\displaystyle \therefore \sin x = - \sqrt{1 - \cos^2 x} = - \sqrt{1 - \frac{9}{25} } = - \frac{4}{5}$ $\displaystyle \therefore \mathrm{cosec} x= \frac{1}{\sin x} = - \frac{5}{4}$ $\displaystyle \tan x = \frac{\sin x }{\cos x} = - \frac{4}{5} \times \frac{-5}{3} = \frac{4}{3}$ $\displaystyle \therefore \cot x = \frac{1}{\tan x} = \frac{3}{4}$ $\displaystyle \therefore \frac{\mathrm{cosec} x + \cot x}{\sec x - \tan x} = \frac{(\frac{-5}{4}) \times \frac{3}{4}}{(\frac{-5}{3}) - \frac{4}{3}} = \frac{1}{6}$