SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Trigonometric Functions

52 questions · 52 still being checked

EXERCISE 3.2 1–10 (part 2 of 6)

  1. Find the values of other five trigonometric functions in Exercises $\displaystyle 1$ to 5.

    Exercise 1

    cos⁡x=−12,x\displaystyle \cos x=-\frac{1}{2}, x lies in third quadrant.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \sin x=-\frac{\sqrt{3}}{2}, \operatorname{cosec} x=-\frac{2}{\sqrt{3}}, \sec x=-2, \tan x=\sqrt{3}, \cot x=\frac{1}{\sqrt{3}}$
    Pythagorean identity plus the quadrant sign. Use \(\displaystyle \sin^{2}x+\cos^{2}x=1\) to get the second ratio, fix its sign from the quadrant, then read off the rest.In the third quadrant both \(\displaystyle \sin x\) and \(\displaystyle \cos x\) are negative (so \(\displaystyle \tan x\) and \(\displaystyle \cot x\) are positive).Given \(\displaystyle \cos x=-\dfrac{1}{2}\), \[\sin^{2}x=1-\cos^{2}x=1-\frac{1}{4}=\frac{3}{4}\quad\Longrightarrow\quad \sin x=-\frac{\sqrt{3}}{2}\] taking the negative root because \(\displaystyle x\) is in the third quadrant.Hence \[\operatorname{cosec}x=\frac{1}{\sin x}=-\frac{2}{\sqrt{3}},\qquad \sec x=\frac{1}{\cos x}=-2,\] \[\tan x=\frac{\sin x}{\cos x}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3},\qquad \cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}}.\]\(\displaystyle \sin x=-\dfrac{\sqrt{3}}{2},\ \operatorname{cosec}x=-\dfrac{2}{\sqrt{3}},\ \sec x=-2,\ \tan x=\sqrt{3},\ \cot x=\dfrac{1}{\sqrt{3}}\).

  2. Exercise 2

    sin⁡x=35,x\displaystyle \sin x=\frac{3}{5}, x lies in second quadrant.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \operatorname{cosec} x=\frac{5}{3}, \cos x=-\frac{4}{5}, \sec x=-\frac{5}{4}, \tan x=-\frac{3}{4}, \cot x=-\frac{4}{3}$
    Pythagorean identity plus the quadrant sign. In the second quadrant \(\displaystyle \sin x>0\) while \(\displaystyle \cos x<0\), so \(\displaystyle \tan x\) and \(\displaystyle \cot x\) are negative.Given \(\displaystyle \sin x=\dfrac{3}{5}\), \[\cos^{2}x=1-\sin^{2}x=1-\frac{9}{25}=\frac{16}{25}\quad\Longrightarrow\quad \cos x=-\frac{4}{5}\] taking the negative root because \(\displaystyle x\) is in the second quadrant.Hence \[\operatorname{cosec}x=\frac{5}{3},\qquad \sec x=-\frac{5}{4},\] \[\tan x=\frac{\sin x}{\cos x}=\frac{3/5}{-4/5}=-\frac{3}{4},\qquad \cot x=-\frac{4}{3}.\]\(\displaystyle \cos x=-\dfrac{4}{5},\ \operatorname{cosec}x=\dfrac{5}{3},\ \sec x=-\dfrac{5}{4},\ \tan x=-\dfrac{3}{4},\ \cot x=-\dfrac{4}{3}\).

  3. Exercise 3

    cot⁡x=34,x\displaystyle \cot x=\frac{3}{4}, x lies in third quadrant.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \sin x=-\frac{4}{5}, \operatorname{cosec} x=-\frac{5}{4}, \cos x=-\frac{3}{5}, \sec x=-\frac{5}{3}, \tan x=\frac{4}{3}$
    Identity \(\displaystyle 1+\cot^{2}x=\operatorname{cosec}^{2}x\) plus the quadrant sign. In the third quadrant \(\displaystyle \sin x<0\) and \(\displaystyle \cos x<0\).Given \(\displaystyle \cot x=\dfrac{3}{4}\), at once \(\displaystyle \tan x=\dfrac{1}{\cot x}=\dfrac{4}{3}\).\[\operatorname{cosec}^{2}x=1+\cot^{2}x=1+\frac{9}{16}=\frac{25}{16}\quad\Longrightarrow\quad \operatorname{cosec}x=-\frac{5}{4}\] taking the negative root because \(\displaystyle \sin x<0\) in the third quadrant. So \[\sin x=-\frac{4}{5}.\]Then, from \(\displaystyle \cos x=\cot x\cdot\sin x\), \[\cos x=\frac{3}{4}\times\left(-\frac{4}{5}\right)=-\frac{3}{5},\qquad \sec x=-\frac{5}{3}.\]\(\displaystyle \tan x=\dfrac{4}{3},\ \sin x=-\dfrac{4}{5},\ \operatorname{cosec}x=-\dfrac{5}{4},\ \cos x=-\dfrac{3}{5},\ \sec x=-\dfrac{5}{3}\).

  4. Exercise 4

    sec⁡x=135,x\displaystyle \sec x=\frac{13}{5}, x lies in fourth quadrant.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \sin x=-\frac{12}{13}, \operatorname{cosec} x=-\frac{13}{12}, \cos x=\frac{5}{13}, \tan x=-\frac{12}{5}, \cot x=-\frac{5}{12}$
    Pythagorean identity plus the quadrant sign. In the fourth quadrant \(\displaystyle \cos x>0\) and \(\displaystyle \sin x<0\).Given \(\displaystyle \sec x=\dfrac{13}{5}\), so \[\cos x=\frac{1}{\sec x}=\frac{5}{13}.\]\[\sin^{2}x=1-\cos^{2}x=1-\frac{25}{169}=\frac{144}{169}\quad\Longrightarrow\quad \sin x=-\frac{12}{13}\] taking the negative root because \(\displaystyle x\) is in the fourth quadrant.Hence \[\operatorname{cosec}x=-\frac{13}{12},\qquad \tan x=\frac{\sin x}{\cos x}=\frac{-12/13}{5/13}=-\frac{12}{5},\qquad \cot x=-\frac{5}{12}.\]\(\displaystyle \cos x=\dfrac{5}{13},\ \sin x=-\dfrac{12}{13},\ \operatorname{cosec}x=-\dfrac{13}{12},\ \tan x=-\dfrac{12}{5},\ \cot x=-\dfrac{5}{12}\).

  5. Exercise 5

    tan⁡x=−512,x\displaystyle \tan x=-\frac{5}{12}, x lies in second quadrant.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \sin x=\frac{5}{13}, \operatorname{cosec} x=\frac{13}{5}, \cos x=-\frac{12}{13}, \sec x=-\frac{13}{12}, \cot x=-\frac{12}{5}$
    Identity \(\displaystyle 1+\tan^{2}x=\sec^{2}x\) plus the quadrant sign. In the second quadrant \(\displaystyle \cos x<0\) and \(\displaystyle \sin x>0\).Given \(\displaystyle \tan x=-\dfrac{5}{12}\), at once \(\displaystyle \cot x=-\dfrac{12}{5}\).\[\sec^{2}x=1+\tan^{2}x=1+\frac{25}{144}=\frac{169}{144}\quad\Longrightarrow\quad \sec x=-\frac{13}{12}\] taking the negative root because \(\displaystyle \cos x<0\) in the second quadrant. So \[\cos x=-\frac{12}{13}.\]Then, from \(\displaystyle \sin x=\tan x\cdot\cos x\), \[\sin x=\left(-\frac{5}{12}\right)\times\left(-\frac{12}{13}\right)=\frac{5}{13},\qquad \operatorname{cosec}x=\frac{13}{5}.\]\(\displaystyle \cot x=-\dfrac{12}{5},\ \sec x=-\dfrac{13}{12},\ \cos x=-\dfrac{12}{13},\ \sin x=\dfrac{5}{13},\ \operatorname{cosec}x=\dfrac{13}{5}\).
  6. Find the values of the trigonometric functions in Exercises $\displaystyle 6$ to 10.

    Exercise 6

    sin⁡765∘\displaystyle \sin 765^{\circ}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \frac{1}{\sqrt{2}}$
    Periodicity. \(\displaystyle \sin\) repeats every \(\displaystyle 360^{\circ}\): \(\displaystyle \sin(n\cdot 360^{\circ}+\theta)=\sin\theta\) for every integer \(\displaystyle n\).Write \(\displaystyle 765^{\circ}\) in that form: \[765^{\circ}=2\times 360^{\circ}+45^{\circ}.\] Therefore \[\sin 765^{\circ}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}.\]\(\displaystyle \sin 765^{\circ}=\dfrac{1}{\sqrt{2}}\).

  7. Exercise 7

    cosec⁡(−1410∘)\displaystyle \operatorname{cosec}\left(-1410^{\circ}\right)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 2$
    Periodicity. \(\displaystyle \operatorname{cosec}\) repeats every \(\displaystyle 360^{\circ}\), so add a whole number of full turns to bring the angle into \(\displaystyle [0^{\circ},360^{\circ})\).Add \(\displaystyle 4\times 360^{\circ}=1440^{\circ}\): \[-1410^{\circ}+1440^{\circ}=30^{\circ}.\] Therefore \[\operatorname{cosec}(-1410^{\circ})=\operatorname{cosec}30^{\circ}=\frac{1}{\sin 30^{\circ}}=\frac{1}{1/2}=2.\]\(\displaystyle \operatorname{cosec}(-1410^{\circ})=2\).

  8. Exercise 8

    tan⁡19π3\displaystyle \tan \frac{19 \pi}{3}

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \sqrt{3}$
    Periodicity. Subtract whole turns of \(\displaystyle 2\pi\) to bring the angle into \(\displaystyle [0,2\pi)\).\[\frac{19\pi}{3}-6\pi=\frac{19\pi-18\pi}{3}=\frac{\pi}{3}\] so \(\displaystyle \dfrac{19\pi}{3}=3(2\pi)+\dfrac{\pi}{3}\), and therefore \[\tan\frac{19\pi}{3}=\tan\frac{\pi}{3}=\sqrt{3}.\]\(\displaystyle \tan\dfrac{19\pi}{3}=\sqrt{3}\).

  9. Exercise 9

    sin⁡(−11π3)\displaystyle \sin \left(-\frac{11 \pi}{3}\right)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle \frac{\sqrt{3}}{2}$
    Periodicity. \(\displaystyle \sin\) has period \(\displaystyle 2\pi\), so add whole turns until the angle lies in \(\displaystyle [0,2\pi)\).Add \(\displaystyle 4\pi\): \[-\frac{11\pi}{3}+4\pi=\frac{-11\pi+12\pi}{3}=\frac{\pi}{3}.\] Therefore \[\sin\left(-\frac{11\pi}{3}\right)=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}.\]\(\displaystyle \sin\left(-\dfrac{11\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\).

  10. Exercise 10

    cot⁡(−15π4)\displaystyle \cot \left(-\frac{15 \pi}{4}\right)

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 1$
    Periodicity. \(\displaystyle \cot\) repeats every \(\displaystyle 2\pi\) (indeed every \(\displaystyle \pi\)), so add whole turns until the angle lies in \(\displaystyle [0,2\pi)\).Add \(\displaystyle 4\pi\): \[-\frac{15\pi}{4}+4\pi=\frac{-15\pi+16\pi}{4}=\frac{\pi}{4}.\] Therefore \[\cot\left(-\frac{15\pi}{4}\right)=\cot\frac{\pi}{4}=1.\]\(\displaystyle \cot\left(-\dfrac{15\pi}{4}\right)=1\).