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NCERT Exemplar · Class 12 Mathematics Inverse Trigonometric Functions

55 questions · 55 still being checked

EXERCISE 2.3 31–40 (part 4 of 6)

  1. Choose the correct answers from the given four options in each of the Exercises from $\displaystyle 20$ to $\displaystyle 37$ (M.C.Q.).

    Exercise 31

    If sin⁡−1(2a1+a2)+cos⁡−1(1−a21+a2)=tan⁡−1(2x1−x2)\displaystyle \sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\cos ^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right), where a,x∈]0,1\displaystyle \left.a, x \in\right] 0,1, then the value of x\displaystyle x is
    (A)
    0\displaystyle 0 (B) a2\displaystyle \frac{a}{2}
    (C)
    a\displaystyle a (D) 2a1−a2\displaystyle \frac{2 a}{1-a^2}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{2a}{1-a^{2}}\)Put \(\displaystyle a=\tan\theta\) and \(\displaystyle x=\tan\varphi\), with \(\displaystyle \theta,\varphi \in \left(0,\tfrac{\pi}{4}\right)\) since \(\displaystyle a,x\in\,]0,1[\).\[\frac{2a}{1+a^{2}} = \sin 2\theta, \quad \frac{1-a^{2}}{1+a^{2}} = \cos 2\theta, \quad \frac{2x}{1-x^{2}} = \tan 2\varphi \]\[\sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\theta) = 2\theta + 2\theta = 4\theta \quad \left(2\theta \in \left(0,\tfrac{\pi}{2}\right)\right) \]\[\tan^{-1}(\tan 2\varphi) = 2\varphi \]\[2\varphi = 4\theta \;\Rightarrow\; \varphi = 2\theta \]\[x = \tan 2\theta = \frac{2a}{1-a^{2}} \]

  2. Exercise 32

    The value of cot⁡[cos⁡−1(725)]\displaystyle \cot \left[\cos ^{-1}\left(\frac{7}{25}\right)\right] is
    (A)
    2524\displaystyle \frac{25}{24}
    (B)
    257\displaystyle \frac{25}{7}
    (C)
    2425\displaystyle \frac{24}{25}
    (D)
    724\displaystyle \frac{7}{24}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{7}{24}\)Let \(\displaystyle \theta = \cos^{-1}\tfrac{7}{25}\), so \(\displaystyle \theta \in \left(0,\tfrac{\pi}{2}\right)\) and \(\displaystyle \sin\theta > 0\).\[\cos\theta = \tfrac{7}{25} \]\[\sin\theta = \sqrt{1-\tfrac{49}{625}} = \tfrac{24}{25} \]\[\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{7}{24} \]

  3. Exercise 33

    The value of the expression tan⁡(12cos⁡−125)\displaystyle \tan \left(\frac{1}{2} \cos ^{-1} \frac{2}{\sqrt{5}}\right) is
    (A)
    2+5\displaystyle 2+\sqrt{5}
    (B)
    5−2\displaystyle \sqrt{5}-2
    (C)
    5+22\displaystyle \frac{\sqrt{5}+2}{2}
    (D)
    5+2\displaystyle 5+\sqrt{2}
    [\displaystyle \left[\right. Hint :tan⁡θ2=1−cos⁡θ1+cos⁡θ]\displaystyle \left.: \tan \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right]

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    NCERT’s answer
    B
    (B) \(\displaystyle \sqrt{5}-2\)Let \(\displaystyle \theta=\cos^{-1}\frac{2}{\sqrt5}\), so \(\displaystyle \cos\theta=\frac{2}{\sqrt5}\) and \(\displaystyle \theta\in\left[0,\frac{\pi}{2}\right]\).\[\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\sqrt{\frac{\sqrt5-2}{\sqrt5+2}} \]\[=\sqrt{\frac{(\sqrt5-2)^2}{(\sqrt5+2)(\sqrt5-2)}}=\sqrt{\frac{(\sqrt5-2)^2}{5-4}} \]\[=\sqrt5-2 \quad (\sqrt5-2>0) \]

  4. Exercise 34

    If ∣x∣≤1\displaystyle |x| \leq 1, then 2tan⁡−1x+sin⁡−1(2x1+x2)\displaystyle 2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) is equal to
    (A)
    4tan⁡−1x\displaystyle 4 \tan ^{-1} x
    (B)
    0\displaystyle 0 (C) π2\displaystyle \frac{\pi}{2}
    (D)
    π\displaystyle \pi

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    NCERT’s answer
    A
    (A) \(\displaystyle 4\tan^{-1}x\)Put \(\displaystyle \varphi=\tan^{-1}x\). For \(\displaystyle |x|\le1\), \(\displaystyle \varphi\in\left[-\frac{\pi}{4},\frac{\pi}{4}\right]\).\[\frac{2x}{1+x^2}=\frac{2\tan\varphi}{1+\tan^2\varphi}=\sin2\varphi \]\[2\varphi\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\ \Rightarrow\ \sin^{-1}(\sin2\varphi)=2\varphi=2\tan^{-1}x \]\[2\tan^{-1}x+2\tan^{-1}x=4\tan^{-1}x \]

  5. Exercise 35

    If cos⁡−1α+cos⁡−1β+cos⁡−1γ=3π\displaystyle \cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi, then α(β+γ)+β(γ+α)+γ(α+β)\displaystyle \alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta) equals
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 6\displaystyle 6 (D) 12\displaystyle 12

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    NCERT’s answer
    C
    (C) \(\displaystyle 6\)Each \(\displaystyle \cos^{-1}\) is at most \(\displaystyle \pi\), so a total of \(\displaystyle 3\pi\) forces each to equal \(\displaystyle \pi\).\[\cos^{-1}\alpha=\cos^{-1}\beta=\cos^{-1}\gamma=\pi\ \Rightarrow\ \alpha=\beta=\gamma=-1 \]\[\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)=3\,(-1)(-2) \]\[=6 \]

  6. Exercise 36

    The number of real solutions of the equation 1+cos⁡2x=2cos⁡−1(cos⁡x)\displaystyle \sqrt{1+\cos 2 x}=\sqrt{2} \cos ^{-1}(\cos x) in [π2,π]\displaystyle \left[\frac{\pi}{2}, \pi\right] is
    (A)
    0\displaystyle 0 (B) 1\displaystyle 1 (C) 2\displaystyle 2 (D) Infinite

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    NCERT’s answer
    A
    (A) \(\displaystyle 0\)On \(\displaystyle \left[\frac{\pi}{2},\pi\right]\): \(\displaystyle \cos x\le0\) and \(\displaystyle \cos^{-1}(\cos x)=x\).\[\sqrt{1+\cos2x}=\sqrt{2\cos^2x}=\sqrt2\,|\cos x|=-\sqrt2\cos x \]\[-\sqrt2\cos x=\sqrt2\,x\ \Rightarrow\ x=-\cos x \]\[-\cos x\le1<\frac{\pi}{2}\le x \]The two sides never meet, so there is no solution.

  7. Exercise 37

    If cos⁡−1x>sin⁡−1x\displaystyle \cos ^{-1} x>\sin ^{-1} x, then
    (A)
    12<x≤1\displaystyle \frac{1}{\sqrt{2}}<x \leq 1
    (B)
    0≤x<12\displaystyle 0 \leq x<\frac{1}{\sqrt{2}}
    (C)
    −1≤x<12\displaystyle -1 \leq x<\frac{1}{\sqrt{2}}
    (D)
    x>0\displaystyle x>0

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    (C) \(\displaystyle -1\le x<\frac{1}{\sqrt2}\)\[\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x \]\[\frac{\pi}{2}-\sin^{-1}x>\sin^{-1}x\ \Rightarrow\ \sin^{-1}x<\frac{\pi}{4} \]Sine is increasing on \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), and the domain gives \(\displaystyle x\ge-1\).\[-1\le x<\sin\frac{\pi}{4}=\frac{1}{\sqrt2} \]
  8. Fill in the blanks in each of the Exercises $\displaystyle 38$ to 48.

    Exercise 38

    The principal value of cos⁡−1(−12)\displaystyle \cos ^{-1}\left(-\frac{1}{2}\right) is ____.

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    NCERT’s answer
    \(\displaystyle \frac{2 \pi}{3}\)
    \(\displaystyle \frac{2\pi}{3}\)\[\cos^{-1}(-x)=\pi-\cos^{-1}x \]\[\cos^{-1}\left(-\frac12\right)=\pi-\frac{\pi}{3}=\frac{2\pi}{3} \]\[\frac{2\pi}{3}\in[0,\pi],\quad \cos\frac{2\pi}{3}=-\frac12 \]Answer: \(\displaystyle \frac{2\pi}{3}\)

  9. Exercise 39

    The value of sin⁡−1(sin⁡3π5)\displaystyle \sin ^{-1}\left(\sin \frac{3 \pi}{5}\right) is ____.

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    NCERT’s answer
    \(\displaystyle \frac{2 \pi}{5}\)
    \(\displaystyle \frac{2\pi}{5}\)\(\displaystyle \frac{3\pi}{5}\) lies outside \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), so reduce it first.\[\sin\frac{3\pi}{5}=\sin\left(\pi-\frac{3\pi}{5}\right)=\sin\frac{2\pi}{5} \]\[\sin^{-1}\left(\sin\frac{2\pi}{5}\right)=\frac{2\pi}{5},\quad \frac{2\pi}{5}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \]Answer: \(\displaystyle \frac{2\pi}{5}\)

  10. Exercise 40

    If cos⁡(tan⁡−1x+cot⁡−13)=0\displaystyle \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0, then value of x\displaystyle x is ____.

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    NCERT’s answer
    \(\displaystyle \sqrt{3}\)
    \(\displaystyle \sqrt3\)\[\cot^{-1}\sqrt3=\frac{\pi}{6} \]\[\cos\left(\tan^{-1}x+\frac{\pi}{6}\right)=0 \]Here \(\displaystyle \tan^{-1}x+\frac{\pi}{6}\in\left(-\frac{\pi}{3},\frac{2\pi}{3}\right)\), where cosine vanishes only at \(\displaystyle \frac{\pi}{2}\).\[\tan^{-1}x+\frac{\pi}{6}=\frac{\pi}{2}\ \Rightarrow\ \tan^{-1}x=\frac{\pi}{3} \]\[x=\tan\frac{\pi}{3}=\sqrt3 \]Answer: \(\displaystyle x=\sqrt3\)