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NCERT Exemplar · Class 12 Mathematics Inverse Trigonometric Functions

55 questions · 55 still being checked

EXERCISE 2.3 41–50 (part 5 of 6)

  1. Fill in the blanks in each of the Exercises $\displaystyle 38$ to 48.

    Exercise 41

    The set of values of sec⁡−1(12)\displaystyle \sec ^{-1}\left(\frac{1}{2}\right) is ____.

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    NCERT’s answer
    \(\displaystyle \phi\)
    \(\displaystyle \varnothing\) (the empty set), since \(\displaystyle \sec^{-1}\) is defined only for \(\displaystyle |y|\ge 1\).\[\sec\theta=\frac{1}{\cos\theta},\qquad |\cos\theta|\le 1 \] \[|\sec\theta|=\frac{1}{|\cos\theta|}\ge 1 \quad\text{for every } \theta \] \[\left|\tfrac12\right|<1 \;\Rightarrow\; \sec\theta=\tfrac12 \text{ has no solution} \] \[\sec^{-1}\!\left(\tfrac12\right)\text{ has no value} \]

  2. Exercise 42

    The principal value of tan⁡−13\displaystyle \tan ^{-1} \sqrt{3} is ____.

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    NCERT’s answer
    \(\displaystyle \frac{\pi}{3}\)
    \(\displaystyle \dfrac{\pi}{3}\)\[\tan\frac{\pi}{3}=\sqrt3 \] \[\frac{\pi}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \quad\text{(principal range of } \tan^{-1}) \] \[\tan^{-1}\sqrt3=\frac{\pi}{3} \]

  3. Exercise 43

    The value of cos⁡−1(cos⁡14π3)\displaystyle \cos ^{-1}\left(\cos \frac{14 \pi}{3}\right) is ____.

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    NCERT’s answer
    \(\displaystyle \frac{2 \pi}{3}\)
    \(\displaystyle \dfrac{2\pi}{3}\)\[\frac{14\pi}{3}=4\pi+\frac{2\pi}{3} \] \[\cos\frac{14\pi}{3}=\cos\left(4\pi+\frac{2\pi}{3}\right)=\cos\frac{2\pi}{3} \] \[\frac{2\pi}{3}\in[0,\pi] \quad\text{(principal range of } \cos^{-1}) \] \[\cos^{-1}\left(\cos\frac{14\pi}{3}\right)=\cos^{-1}\left(\cos\frac{2\pi}{3}\right)=\frac{2\pi}{3} \]

  4. Exercise 44

    The value of cos⁡(sin⁡−1x+cos⁡−1x),∣x∣≤1\displaystyle \cos \left(\sin ^{-1} x+\cos ^{-1} x\right),|x| \leq 1 is ____.

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    NCERT’s answer
    $\displaystyle 0$
    \(\displaystyle 0\)\[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\qquad |x|\le 1 \] \[\cos\left(\sin^{-1}x+\cos^{-1}x\right)=\cos\frac{\pi}{2}=0 \]

  5. Exercise 45

    The value of expression tan⁡(sin⁡−1x+cos⁡−1x2)\displaystyle \tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right),when x=32\displaystyle x=\frac{\sqrt{3}}{2} is ____.

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    NCERT’s answer
    $\displaystyle 1$
    \(\displaystyle 1\)\[x=\frac{\sqrt3}{2}\in[-1,1],\qquad \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2} \] \[\tan\left(\frac{\sin^{-1}x+\cos^{-1}x}{2}\right)=\tan\frac{\pi}{4}=1 \]

  6. Exercise 46

    If y=2tan⁡−1x+sin⁡−1(2x1+x2)\displaystyle y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) for all x\displaystyle x, then ____ <y<\displaystyle <y< ____.

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    NCERT’s answer
    \(\displaystyle -2 \pi, 2 \pi\)
    \(\displaystyle -2\pi\) and \(\displaystyle 2\pi\), i.e. \(\displaystyle -2\pi<y<2\pi\).Put \(\displaystyle x=\tan\theta\), \(\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\):\[\frac{2x}{1+x^2}=\sin2\theta,\qquad y=2\theta+\sin^{-1}(\sin2\theta) \] \[|x|\le1:\; 2\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\Rightarrow \sin^{-1}(\sin2\theta)=2\theta,\quad y=4\tan^{-1}x\in[-\pi,\pi] \] \[x>1:\; 2\theta\in\left(\tfrac{\pi}{2},\pi\right)\Rightarrow \sin^{-1}(\sin2\theta)=\pi-2\theta,\quad y=2\theta+\pi-2\theta=\pi \] \[x<-1:\; 2\theta\in\left(-\pi,-\tfrac{\pi}{2}\right)\Rightarrow \sin^{-1}(\sin2\theta)=-\pi-2\theta,\quad y=-\pi \] \[-\pi\le y\le\pi\ \Rightarrow\ -2\pi<y<2\pi \]\(\displaystyle y\) equals \(\displaystyle \pm\pi\) for \(\displaystyle |x|\ge1\), so strict bounds must lie beyond \(\displaystyle \pm\pi\).

  7. Exercise 47

    The result tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\displaystyle \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right) is true when value of xy\displaystyle x y is ____

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    NCERT’s answer
    \(\displaystyle x y>-1\)
    \(\displaystyle xy>-1\)\[\tan^{-1}x-\tan^{-1}y=\tan^{-1}x+\tan^{-1}(-y) \] \[\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}\quad\text{only when } ab<1 \] \[a=x,\; b=-y:\quad \tan^{-1}x+\tan^{-1}(-y)=\tan^{-1}\frac{x-y}{1+xy}\quad\text{only when } -xy<1 \] \[-xy<1\;\Longleftrightarrow\; xy>-1 \]

  8. Exercise 48

    The value of cot⁡−1(−x)\displaystyle \cot ^{-1}(-x) for all x∈R\displaystyle x \in \mathbf{R} in terms of cot⁡−1x\displaystyle \cot ^{-1} x is ____.

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    NCERT’s answer
    \(\displaystyle \pi-\cot ^{-1} x\)
    \(\displaystyle \pi-\cot^{-1}x\)\[\theta=\cot^{-1}x,\quad \theta\in(0,\pi),\quad \cot\theta=x \] \[\cot(\pi-\theta)=-\cot\theta=-x,\qquad \pi-\theta\in(0,\pi) \] \[\cot^{-1}(-x)=\pi-\theta=\pi-\cot^{-1}x \]
  9. State True or False for the statement in each of the Exercises $\displaystyle 49$ to 55.

    Exercise 49

    All trigonometric functions have inverse over their respective domains.

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    NCERT’s answer
    False
    False. A trigonometric function is many-one on its whole domain, so no inverse exists there; it needs a restricted branch.\[\sin 0 = \sin \pi = 0 \quad \text{(two inputs, one output)} \]

  10. Exercise 50

    The value of the expression (cos⁡−1x)2\displaystyle \left(\cos ^{-1} x\right)^2 is equal to sec⁡2x\displaystyle \sec ^2 x.

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    NCERT’s answer
    False
    False. Test with \(\displaystyle x = 0\).\[\left(\cos^{-1} 0\right)^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \]\[\sec^2 0 = 1 \]\[\frac{\pi^2}{4} \neq 1 \]