SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Inverse Trigonometric Functions

43 questions · 43 still being checked

EXERCISE 2.2 11–15 (part 4 of 6)

  1. Find the values of each of the following:

    Exercise 11

    tan⁡−1(tan⁡3π4)\displaystyle \tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)

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    NCERT’s answer
    \(\displaystyle \frac{-\pi}{4}\)
    \(\displaystyle \tan^{-1}(\tan\theta)=\theta\) only for \(\displaystyle \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\), and \(\displaystyle \frac{3\pi}{4}\) is not in that interval, so the answer is not \(\displaystyle \frac{3\pi}{4}\).The tangent has period \(\displaystyle \pi\), so subtract \(\displaystyle \pi\) to bring the angle into the principal branch: \[\tan\frac{3\pi}{4}=\tan\left(\frac{3\pi}{4}-\pi\right)=\tan\left(-\frac{\pi}{4}\right),\qquad -\frac{\pi}{4}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).\]Therefore \[\tan^{-1}\left(\tan\frac{3\pi}{4}\right)=-\frac{\pi}{4}.\]

  2. Exercise 12

    tan⁡(sin⁡−135+cot⁡−132)\displaystyle \tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)

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    NCERT’s answer
    \(\displaystyle \frac{17}{6}\)
    Name the two angles and find their tangents.Let \(\displaystyle A=\sin^{-1}\frac{3}{5}\). Then \(\displaystyle A\in\left(0,\frac{\pi}{2}\right)\) with \(\displaystyle \sin A=\frac{3}{5}\), so \(\displaystyle \cos A=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\) (positive, as \(\displaystyle A\) is in the first quadrant) and \[\tan A=\frac{3}{4}.\]Let \(\displaystyle B=\cot^{-1}\frac{3}{2}\), so \(\displaystyle \cot B=\frac{3}{2}\) and \(\displaystyle \tan B=\frac{2}{3}\).Apply the addition formula: \[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\dfrac{3}{4}+\dfrac{2}{3}}{1-\dfrac{3}{4}\cdot\dfrac{2}{3}}=\frac{\dfrac{17}{12}}{\dfrac{1}{2}}=\frac{17}{6}.\]\[\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\right)=\frac{17}{6}.\]

  3. Exercise 13

    cos⁡−1(cos⁡7π6)\displaystyle \cos ^{-1}\left(\cos \frac{7 \pi}{6}\right) is equal to (A) 7π6\displaystyle \frac{7 \pi}{6} (B) 5π6\displaystyle \frac{5 \pi}{6} (C) π3\displaystyle \frac{\pi}{3} (D) π6\displaystyle \frac{\pi}{6}

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    NCERT’s answer
    B
    \(\displaystyle \cos^{-1}(\cos\theta)=\theta\) only for \(\displaystyle \theta\in[0,\pi]\), the principal branch. Here \(\displaystyle \frac{7\pi}{6}>\pi\), so option (A) is the trap; convert to an angle in \(\displaystyle [0,\pi]\) with the same cosine.Using \(\displaystyle \cos(2\pi-\theta)=\cos\theta\), \[\cos\frac{7\pi}{6}=\cos\left(2\pi-\frac{7\pi}{6}\right)=\cos\frac{5\pi}{6},\qquad \frac{5\pi}{6}\in[0,\pi].\]Therefore \(\displaystyle \cos^{-1}\left(\cos\frac{7\pi}{6}\right)=\frac{5\pi}{6}\).Answer: (B) \(\displaystyle \frac{5\pi}{6}\).

  4. Exercise 14

    sin⁡(π3−sin⁡−1(−12))\displaystyle \sin \left(\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right) is equal to (A) 12\displaystyle \frac{1}{2} (B) 13\displaystyle \frac{1}{3} (C) 14\displaystyle \frac{1}{4} (D) 1\displaystyle 1

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    NCERT’s answer
    D
    First evaluate the inverse function as a principal value. \(\displaystyle \sin^{-1}\) takes values in \(\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]\), and \(\displaystyle \sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}\), so \[\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}.\]Watch the double negative in the subtraction: \[\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2}.\]Hence the expression is \(\displaystyle \sin\frac{\pi}{2}=1\).Answer: (D) \(\displaystyle 1\).

  5. Exercise 15

    tan⁡−13−cot⁡−1(−3)\displaystyle \tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3}) is equal to (A) π\displaystyle \pi (B) −π2\displaystyle -\frac{\pi}{2} (C) 0\displaystyle 0 (D) 23\displaystyle 2 \sqrt{3}

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    NCERT’s answer
    B
    \(\displaystyle \tan^{-1}\sqrt{3}=\frac{\pi}{3}\), since \(\displaystyle \frac{\pi}{3}\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) and \(\displaystyle \tan\frac{\pi}{3}=\sqrt{3}\).For the second term, the principal branch of \(\displaystyle \cot^{-1}\) is \(\displaystyle (0,\pi)\), so \(\displaystyle \cot^{-1}\) of a negative number is an obtuse angle, not a negative one. Using \(\displaystyle \cot^{-1}(-t)=\pi-\cot^{-1}t\), \[\cot^{-1}\left(-\sqrt{3}\right)=\pi-\cot^{-1}\sqrt{3}=\pi-\frac{\pi}{6}=\frac{5\pi}{6},\] and indeed \(\displaystyle \cot\frac{5\pi}{6}=-\sqrt{3}\).Therefore \[\tan^{-1}\sqrt{3}-\cot^{-1}\left(-\sqrt{3}\right)=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{\pi}{2}.\]Answer: (B) \(\displaystyle -\frac{\pi}{2}\).