Class 11: Trigonometric Functions

$\displaystyle \textbf{Question 1: }\text{Find the values of the other five trigonometric functions in each of the}$
$\displaystyle \text{following:}$
$\displaystyle \text{(i) }\cot x=\frac{12}{5},\ x\text{ is in Quadrant III}$
$\displaystyle \text{(ii) }\cos x=-\frac{1}{2},\ x\text{ is in Quadrant II}$
$\displaystyle \text{(iii) }\tan x=\frac{3}{4},\ x\text{ is in Quadrant III}$
$\displaystyle \text{(iv) }\sin x=\frac{3}{5},\ x\text{ is in Quadrant I}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }\cot x=\frac{12}{5},\ x\text{ is in Quadrant III.}$
$\displaystyle \mathrm{cosec}^2x-\cot^2x=1$
$\displaystyle \therefore \mathrm{cosec}\,x=\pm\sqrt{1+\cot^2x}$
$\displaystyle \text{In Quadrant III, }\mathrm{cosec}\,x<0.$
$\displaystyle \therefore \mathrm{cosec}\,x=-\sqrt{1+\left(\frac{12}{5}\right)^2}$
$\displaystyle =-\sqrt{\frac{169}{25}}=-\frac{13}{5}$
$\displaystyle \tan x=\frac{1}{\cot x}=\frac{5}{12}$
$\displaystyle \sin x=\frac{1}{\mathrm{cosec}\,x}=-\frac{5}{13}$
$\displaystyle \text{In Quadrant III, }\cos x<0.$
$\displaystyle \therefore \cos x=-\sqrt{1-\sin^2x}$
$\displaystyle =-\sqrt{1-\left(-\frac{5}{13}\right)^2}=-\frac{12}{13}$
$\displaystyle \sec x=\frac{1}{\cos x}=-\frac{13}{12}$
$\displaystyle \text{Thus, }\sin x=-\frac{5}{13},\ \cos x=-\frac{12}{13},\ \tan x=\frac{5}{12},$
$\displaystyle \mathrm{cosec}\,x=-\frac{13}{5}\text{ and }\sec x=-\frac{13}{12}.$
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$\displaystyle \text{(ii) Given, }\cos x=-\frac{1}{2},\ x\text{ is in Quadrant II.}$
$\displaystyle \sin^2x+\cos^2x=1$
$\displaystyle \therefore \sin x=\pm\sqrt{1-\cos^2x}$
$\displaystyle \text{In Quadrant II, }\sin x>0.$
$\displaystyle \therefore \sin x=\sqrt{1-\left(-\frac{1}{2}\right)^2}=\frac{\sqrt3}{2}$
$\displaystyle \mathrm{cosec}\,x=\frac{1}{\sin x}=\frac{2}{\sqrt3}=\frac{2\sqrt3}{3}$
$\displaystyle \tan x=\frac{\sin x}{\cos x}=\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}}=-\sqrt3$
$\displaystyle \cot x=\frac{1}{\tan x}=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3}$
$\displaystyle \sec x=\frac{1}{\cos x}=-2$
$\displaystyle \text{Thus, }\sin x=\frac{\sqrt3}{2},\ \tan x=-\sqrt3,\ \cot x=-\frac{\sqrt3}{3},$
$\displaystyle \mathrm{cosec}\,x=\frac{2\sqrt3}{3}\text{ and }\sec x=-2.$
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$\displaystyle \text{(iii) Given, }\tan x=\frac{3}{4},\ x\text{ is in Quadrant III.}$
$\displaystyle \cot x=\frac{1}{\tan x}=\frac{4}{3}$
$\displaystyle \mathrm{cosec}^2x-\cot^2x=1$
$\displaystyle \therefore \mathrm{cosec}\,x=\pm\sqrt{1+\cot^2x}$
$\displaystyle \text{In Quadrant III, }\mathrm{cosec}\,x<0.$
$\displaystyle \therefore \mathrm{cosec}\,x=-\sqrt{1+\left(\frac{4}{3}\right)^2}=-\frac{5}{3}$
$\displaystyle \sin x=\frac{1}{\mathrm{cosec}\,x}=-\frac{3}{5}$
$\displaystyle \tan x=\frac{\sin x}{\cos x}$
$\displaystyle \therefore \cos x=\frac{\sin x}{\tan x}=\frac{-\frac{3}{5}}{\frac{3}{4}}=-\frac{4}{5}$
$\displaystyle \sec x=\frac{1}{\cos x}=-\frac{5}{4}$
$\displaystyle \text{Thus, }\sin x=-\frac{3}{5},\ \cos x=-\frac{4}{5},\ \cot x=\frac{4}{3},$
$\displaystyle \mathrm{cosec}\,x=-\frac{5}{3}\text{ and }\sec x=-\frac{5}{4}.$
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$\displaystyle \text{(iv) Given, }\sin x=\frac{3}{5},\ x\text{ is in Quadrant I.}$
$\displaystyle \text{In Quadrant I, }\cos x>0.$
$\displaystyle \therefore \cos x=\sqrt{1-\sin^2x}$
$\displaystyle =\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}$
$\displaystyle \tan x=\frac{\sin x}{\cos x}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
$\displaystyle \cot x=\frac{1}{\tan x}=\frac{4}{3}$
$\displaystyle \mathrm{cosec}\,x=\frac{1}{\sin x}=\frac{5}{3}$
$\displaystyle \sec x=\frac{1}{\cos x}=\frac{5}{4}$
$\displaystyle \text{Thus, }\cos x=\frac{4}{5},\ \tan x=\frac{3}{4},\ \cot x=\frac{4}{3},$
$\displaystyle \mathrm{cosec}\,x=\frac{5}{3}\text{ and }\sec x=\frac{5}{4}.$
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$\displaystyle \textbf{Question 2: }\text{If }\sin x=\frac{12}{13}\text{ and }x\text{ lies in Quadrant II, find the value of}$
$\displaystyle \sec x+\tan x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin x=\frac{12}{13}\text{ and }x\text{ lies in Quadrant II.}$
$\displaystyle \text{In Quadrant II, }\cos x<0.$
$\displaystyle \therefore \cos x=-\sqrt{1-\sin^2x}$
$\displaystyle =-\sqrt{1-\left(\frac{12}{13}\right)^2}$
$\displaystyle =-\sqrt{\frac{25}{169}}=-\frac{5}{13}$
$\displaystyle \therefore \sec x=\frac{1}{\cos x}=-\frac{13}{5}$
$\displaystyle \tan x=\frac{\sin x}{\cos x}$
$\displaystyle =\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{5}$
$\displaystyle \therefore \sec x+\tan x=-\frac{13}{5}-\frac{12}{5}$
$\displaystyle =-\frac{25}{5}=-5$
$\displaystyle \therefore \text{The required value is }-5.$
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$\displaystyle \textbf{Question 3: }\text{If }\sin x=\frac{3}{5},\ \tan y=\frac{1}{2},\ \frac{\pi}{2}<x<\pi\text{ and}$
$\displaystyle \pi<y<\frac{3\pi}{2},\text{ find the value of }8\tan x-\sqrt5\sec y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin x=\frac{3}{5},\ \tan y=\frac{1}{2},\ \frac{\pi}{2}<x<\pi$
$\displaystyle \text{and }\pi<y<\frac{3\pi}{2}.$
$\displaystyle \therefore x\text{ lies in Quadrant II and }y\text{ lies in Quadrant III.}$
$\displaystyle \text{In Quadrant II, }\cos x<0.$
$\displaystyle \therefore \cos x=-\sqrt{1-\sin^2x}$
$\displaystyle =-\sqrt{1-\left(\frac{3}{5}\right)^2}$
$\displaystyle =-\sqrt{\frac{16}{25}}=-\frac{4}{5}$
$\displaystyle \therefore \tan x=\frac{\sin x}{\cos x}$
$\displaystyle =\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}$
$\displaystyle \sec^2y=1+\tan^2y$
$\displaystyle =1+\left(\frac{1}{2}\right)^2=\frac{5}{4}$
$\displaystyle \text{In Quadrant III, }\sec y<0.$
$\displaystyle \therefore \sec y=-\sqrt{\frac{5}{4}}=-\frac{\sqrt5}{2}$
$\displaystyle \therefore 8\tan x-\sqrt5\sec y$
$\displaystyle =8\left(-\frac{3}{4}\right)-\sqrt5\left(-\frac{\sqrt5}{2}\right)$
$\displaystyle =-6+\frac{5}{2}=-\frac{7}{2}$
$\displaystyle \therefore \text{The required value is }-\frac{7}{2}.$
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$\displaystyle \textbf{Question 4: }\text{If }\sin x+\cos x=0\text{ and }x\text{ lies in the fourth quadrant, find}$
$\displaystyle \sin x\text{ and }\cos x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\sin x+\cos x=0\text{ and }x\text{ lies in Quadrant IV.}$
$\displaystyle \therefore \sin x=-\cos x$
$\displaystyle \therefore \frac{\sin x}{\cos x}=-1$
$\displaystyle \therefore \tan x=-1$
$\displaystyle \sec^2x=1+\tan^2x$
$\displaystyle =1+(-1)^2=2$
$\displaystyle \text{In Quadrant IV, }\sec x>0.$
$\displaystyle \therefore \sec x=\sqrt2$
$\displaystyle \therefore \cos x=\frac{1}{\sec x}=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}$
$\displaystyle \sin x=\tan x\cos x$
$\displaystyle =(-1)\left(\frac{\sqrt2}{2}\right)=-\frac{\sqrt2}{2}$
$\displaystyle \therefore \sin x=-\frac{\sqrt2}{2}\text{ and }\cos x=\frac{\sqrt2}{2}.$
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$\displaystyle \textbf{Question 5: }\text{If }\cos x=-\frac{3}{5}\text{ and }\pi<x<\frac{3\pi}{2},\text{ find the values of the other}$
$\displaystyle \text{five trigonometric functions and hence evaluate }\frac{\mathrm{cosec}\,x+\cot x}{\sec x-\tan x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\cos x=-\frac{3}{5}\text{ and }\pi<x<\frac{3\pi}{2}.$
$\displaystyle \therefore x\text{ lies in Quadrant III.}$
$\displaystyle \sec x=\frac{1}{\cos x}=-\frac{5}{3}$
$\displaystyle \text{In Quadrant III, }\sin x<0.$
$\displaystyle \therefore \sin x=-\sqrt{1-\cos^2x}$
$\displaystyle =-\sqrt{1-\left(-\frac{3}{5}\right)^2}$
$\displaystyle =-\sqrt{\frac{16}{25}}=-\frac{4}{5}$
$\displaystyle \mathrm{cosec}\,x=\frac{1}{\sin x}=-\frac{5}{4}$
$\displaystyle \tan x=\frac{\sin x}{\cos x}$
$\displaystyle =\frac{-\frac{4}{5}}{-\frac{3}{5}}=\frac{4}{3}$
$\displaystyle \cot x=\frac{1}{\tan x}=\frac{3}{4}$
$\displaystyle \therefore \sin x=-\frac{4}{5},\ \tan x=\frac{4}{3},\ \cot x=\frac{3}{4},$
$\displaystyle \mathrm{cosec}\,x=-\frac{5}{4}\text{ and }\sec x=-\frac{5}{3}.$
$\displaystyle \text{Now, }\frac{\mathrm{cosec}\,x+\cot x}{\sec x-\tan x}$
$\displaystyle =\frac{-\frac{5}{4}+\frac{3}{4}}{-\frac{5}{3}-\frac{4}{3}}$
$\displaystyle =\frac{-\frac{2}{4}}{-\frac{9}{3}}$
$\displaystyle =\frac{-\frac{1}{2}}{-3}=\frac{1}{6}$
$\displaystyle \therefore \text{The required value is }\frac{1}{6}.$
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