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NCERT Exemplar · Class 12 Mathematics Inverse Trigonometric Functions

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EXERCISE 2.3 1–10 (part 1 of 6)

  1. Exercise 1

    Find the value of tan⁡−1(tan⁡5π6)+cos⁡−1(cos⁡13π6)\displaystyle \tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right).

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    NCERT’s answer
    $\displaystyle 0$
    Principal values: \[\tan^{-1}x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right),\qquad \cos^{-1}x\in[0,\pi] \] \[\tan\frac{5\pi}{6}=\tan\left(\pi-\frac{\pi}{6}\right)=-\frac{1}{\sqrt3} \] \[\tan^{-1}\left(-\frac{1}{\sqrt3}\right)=-\frac{\pi}{6} \] \[\cos\frac{13\pi}{6}=\cos\left(2\pi+\frac{\pi}{6}\right)=\frac{\sqrt3}{2} \] \[\cos^{-1}\frac{\sqrt3}{2}=\frac{\pi}{6} \] \[-\frac{\pi}{6}+\frac{\pi}{6}=0 \] Answer: \(\displaystyle 0\)

  2. Exercise 2

    Evaluate cos⁡[cos⁡−1(−32)+π6]\displaystyle \cos \left[\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\frac{\pi}{6}\right].

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    NCERT’s answer
    \(\displaystyle -1\)
    \[\cos^{-1}\left(-\frac{\sqrt3}{2}\right)=\pi-\cos^{-1}\frac{\sqrt3}{2}=\pi-\frac{\pi}{6}=\frac{5\pi}{6} \] \[\cos\left[\frac{5\pi}{6}+\frac{\pi}{6}\right]=\cos\pi=-1 \] Answer: \(\displaystyle -1\)

  3. Exercise 3

    Prove that cot⁡(π4−2cot⁡−13)=7\displaystyle \cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)=7.

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    Let \(\displaystyle \theta=\cot^{-1}3\), so \(\displaystyle \cot\theta=3\). \[\tan\theta=\frac13 \] \[\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=\frac{2/3}{1-1/9}=\frac34 \] \[\tan\left(\frac{\pi}{4}-2\theta\right)=\frac{1-\tan2\theta}{1+\tan2\theta}=\frac{1-3/4}{1+3/4}=\frac17 \] \[\cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)=\frac{1}{\tan\left(\dfrac{\pi}{4}-2\theta\right)}=7 \] Answer: LHS \(\displaystyle =7\)

  4. Exercise 4

    Find the value of tan⁡−1(−13)+cot⁡−1(13)+tan⁡−1(sin⁡(−π2))\displaystyle \tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right).

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    NCERT’s answer
    \(\displaystyle \frac{-\pi}{12}\)
    \[\tan^{-1}\left(-\frac{1}{\sqrt3}\right)=-\frac{\pi}{6} \] \[\cot^{-1}\frac{1}{\sqrt3}=\frac{\pi}{3}\qquad(\cot^{-1}x\in(0,\pi)) \] \[\tan^{-1}\left(\sin\left(-\frac{\pi}{2}\right)\right)=\tan^{-1}(-1)=-\frac{\pi}{4} \] \[-\frac{\pi}{6}+\frac{\pi}{3}-\frac{\pi}{4}=\frac{-2\pi+4\pi-3\pi}{12}=-\frac{\pi}{12} \] Answer: \(\displaystyle -\dfrac{\pi}{12}\)

  5. Exercise 5

    Find the value of tan⁡−1(tan⁡2π3)\displaystyle \tan ^{-1}\left(\tan \frac{2 \pi}{3}\right).

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    NCERT’s answer
    \(\displaystyle -\frac{\pi}{3}\)
    \(\displaystyle \dfrac{2\pi}{3}\) lies outside \(\displaystyle \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\), so the value is not \(\displaystyle \dfrac{2\pi}{3}\). \[\tan\frac{2\pi}{3}=\tan\left(\pi-\frac{\pi}{3}\right)=-\sqrt3 \] \[\tan^{-1}\left(-\sqrt3\right)=-\frac{\pi}{3} \] Answer: \(\displaystyle -\dfrac{\pi}{3}\)

  6. Exercise 6

    Show that 2tan⁡−1(−3)=−π2+tan⁡−1(−43)\displaystyle 2 \tan ^{-1}(-3)=\frac{-\pi}{2}+\tan ^{-1}\left(\frac{-4}{3}\right).

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    Let \(\displaystyle \alpha=\tan^{-1}3\). Since \(\displaystyle 3>1\), \(\displaystyle \alpha\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)\), so \(\displaystyle 2\alpha\in\left(\frac{\pi}{2},\pi\right)\). \[\tan2\alpha=\frac{2\cdot3}{1-3^2}=-\frac34 \] Since \(\displaystyle 2\alpha\) is in the second quadrant: \[2\alpha=\pi+\tan^{-1}\left(-\frac34\right)=\pi-\tan^{-1}\frac34 \] \[\text{LHS}=2\tan^{-1}(-3)=-2\alpha=-\pi+\tan^{-1}\frac34 \] \[\tan^{-1}\frac43=\frac{\pi}{2}-\tan^{-1}\frac34 \qquad\left(\tan^{-1}x+\tan^{-1}\tfrac1x=\tfrac{\pi}{2},\ x>0\right) \] \[\text{RHS}=-\frac{\pi}{2}-\tan^{-1}\frac43=-\frac{\pi}{2}-\frac{\pi}{2}+\tan^{-1}\frac34=-\pi+\tan^{-1}\frac34 \] Answer: LHS \(\displaystyle =\) RHS \(\displaystyle =-\pi+\tan^{-1}\dfrac34\)

  7. Exercise 7

    Find the real solutions of the equation tan⁡−1x(x+1)+sin⁡−1x2+x+1=π2.\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2} .

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    NCERT’s answer
    \(\displaystyle 0, -1\)
    Both inverse functions must be defined: \[x(x+1)\ge0 \qquad\text{and}\qquad x^2+x+1\le1 \] The second condition gives \(\displaystyle x(x+1)\le0\), so \[x(x+1)=0 \;\Rightarrow\; x=0 \ \text{or}\ x=-1 \] Check: \[x=0:\ \tan^{-1}0+\sin^{-1}1=0+\frac{\pi}{2}=\frac{\pi}{2} \] \[x=-1:\ \tan^{-1}0+\sin^{-1}1=0+\frac{\pi}{2}=\frac{\pi}{2} \] Answer: \(\displaystyle x=0,\ -1\)

  8. Exercise 8

    Find the value of the expression sin⁡(2tan⁡−113)+cos⁡(tan⁡−122)\displaystyle \sin \left(2 \tan ^{-1} \frac{1}{3}\right)+\cos \left(\tan ^{-1} 2 \sqrt{2}\right).

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    NCERT’s answer
    \(\displaystyle \frac{14}{15}\)
    Let \(\displaystyle \theta=\tan^{-1}\frac13\) and \(\displaystyle \varphi=\tan^{-1}2\sqrt2\); both lie in \(\displaystyle \left(0,\frac{\pi}{2}\right)\). \[\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2/3}{1+1/9}=\frac35 \] \[\cos\varphi=\frac{1}{\sqrt{1+\tan^2\varphi}}=\frac{1}{\sqrt{1+8}}=\frac13 \] \[\frac35+\frac13=\frac{9+5}{15}=\frac{14}{15} \] Answer: \(\displaystyle \dfrac{14}{15}\)

  9. Exercise 9

    If 2tan⁡−1(cos⁡θ)=tan⁡−1(2cosec⁡θ)\displaystyle 2 \tan ^{-1}(\cos \theta)=\tan ^{-1}(2 \operatorname{cosec} \theta), then show that θ=π4\displaystyle \theta=\frac{\pi}{4}, where n\displaystyle n is any integer.

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    \(\displaystyle \operatorname{cosec}\theta\) exists, so \(\displaystyle \sin\theta\neq0\) and \(\displaystyle |\cos\theta|<1\); the double-angle formula applies. \[2\tan^{-1}(\cos\theta)=\tan^{-1}\frac{2\cos\theta}{1-\cos^{2}\theta}=\tan^{-1}\frac{2\cos\theta}{\sin^{2}\theta} \] Equate with the right side (\(\displaystyle \tan^{-1}\) is one-one): \[\frac{2\cos\theta}{\sin^{2}\theta}=\frac{2}{\sin\theta} \] \[\cos\theta=\sin\theta \] \[\tan\theta=1 \] Answer: \(\displaystyle \theta=n\pi+\dfrac{\pi}{4},\ n\in\mathbb{Z}\) (\(\displaystyle \theta=\dfrac{\pi}{4}\) for \(\displaystyle n=0\)).

  10. Exercise 10

    Show that cos⁡(2tan⁡−117)=sin⁡(4tan⁡−113)\displaystyle \cos \left(2 \tan ^{-1} \frac{1}{7}\right)=\sin \left(4 \tan ^{-1} \frac{1}{3}\right).

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    Left side: \[\cos 2\alpha=\frac{1-\tan^{2}\alpha}{1+\tan^{2}\alpha} \] \[\cos\!\left(2\tan^{-1}\tfrac17\right)=\frac{1-\dfrac{1}{49}}{1+\dfrac{1}{49}}=\frac{24}{25} \] Right side: \[2\tan^{-1}\tfrac13=\tan^{-1}\frac{2\cdot\frac13}{1-\frac19}=\tan^{-1}\tfrac34 \] \[4\tan^{-1}\tfrac13=2\tan^{-1}\tfrac34 \] \[\sin\!\left(2\tan^{-1}\tfrac34\right)=\frac{2\cdot\frac34}{1+\dfrac{9}{16}}=\frac{24}{25} \] Answer: both sides equal \(\displaystyle \dfrac{24}{25}\).