SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Inverse Trigonometric Functions

55 questions · 55 still being checked

EXERCISE 2.3 11–20 (part 2 of 6)

  1. Exercise 11

    Solve the following equation cos⁡(tan⁡−1x)=sin⁡(cot⁡−134)\displaystyle \cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right).

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{-3}{4}, \frac{3}{4}\)
    \[\theta=\cot^{-1}\tfrac34\ \Rightarrow\ \cot\theta=\tfrac34,\quad \sin\theta=\frac{1}{\sqrt{1+\cot^{2}\theta}}=\frac45 \] \[\cos(\tan^{-1}x)=\frac{1}{\sqrt{1+x^{2}}} \] \[\frac{1}{\sqrt{1+x^{2}}}=\frac45 \] \[1+x^{2}=\frac{25}{16} \] \[x^{2}=\frac{9}{16} \] Both signs satisfy the equation, since \(\displaystyle \cos(\tan^{-1}x)>0\) for every \(\displaystyle x\). Answer: \(\displaystyle x=\pm\dfrac34\)

  2. Exercise 12

    Prove that tan⁡−1(1+x2+1−x21+x2−1−x2)=π4+12cos⁡−1x2\displaystyle \tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    The expression needs \(\displaystyle 0<|x|\le1\) (denominator nonzero, \(\displaystyle \sqrt{1-x^{2}}\) real). Put \[x^{2}=\cos2\theta,\quad 2\theta=\cos^{-1}x^{2},\quad 0\le\theta<\tfrac{\pi}{4} \] \[1+x^{2}=2\cos^{2}\theta,\qquad 1-x^{2}=2\sin^{2}\theta \] \[\sqrt{1+x^{2}}=\sqrt2\cos\theta,\qquad \sqrt{1-x^{2}}=\sqrt2\sin\theta \] \[\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=\frac{1+\tan\theta}{1-\tan\theta}=\tan\!\left(\frac{\pi}{4}+\theta\right) \] Here \(\displaystyle \frac{\pi}{4}+\theta\in\left[\frac{\pi}{4},\frac{\pi}{2}\right)\), inside the principal range of \(\displaystyle \tan^{-1}\): \[\tan^{-1}\tan\!\left(\frac{\pi}{4}+\theta\right)=\frac{\pi}{4}+\theta \] Answer: the left side \(\displaystyle =\dfrac{\pi}{4}+\dfrac12\cos^{-1}x^{2}\).

  3. Exercise 13

    Find the simplified form of cos⁡−1(35cos⁡x+45sin⁡x)\displaystyle \cos ^{-1}\left(\frac{3}{5} \cos x+\frac{4}{5} \sin x\right), where x∈[−3π4,π4]\displaystyle x \in\left[\frac{-3 \pi}{4}, \frac{\pi}{4}\right].

    Check this one against your book

    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[\cos\alpha=\tfrac35,\quad \sin\alpha=\tfrac45,\quad \alpha=\tan^{-1}\tfrac43\in\left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right) \] \[\tfrac35\cos x+\tfrac45\sin x=\cos(x-\alpha)=\cos(\alpha-x) \] \(\displaystyle \cos^{-1}(\cos t)=t\) only for \(\displaystyle t\in[0,\pi]\): \[0\le\alpha-x\le\pi\iff \alpha-\pi\le x\le\alpha \] This covers \(\displaystyle \left[\alpha-\pi,\tfrac{\pi}{4}\right]\). On the rest, \(\displaystyle -\tfrac{3\pi}{4}\le x<\alpha-\pi\), use \(\displaystyle t=x-\alpha+2\pi\in[0,\pi]\), which has the same cosine. \[\cos^{-1}\!\left(\tfrac35\cos x+\tfrac45\sin x\right)=\begin{cases}\alpha-x, & \alpha-\pi\le x\le\tfrac{\pi}{4}\\ x-\alpha+2\pi, & -\tfrac{3\pi}{4}\le x<\alpha-\pi\end{cases} \] Answer: \(\displaystyle \tan^{-1}\dfrac43-x\) for \(\displaystyle \tan^{-1}\dfrac43-\pi\le x\le\dfrac{\pi}{4}\); on the remaining strip \(\displaystyle -\dfrac{3\pi}{4}\le x<\tan^{-1}\dfrac43-\pi\) it is \(\displaystyle x+2\pi-\tan^{-1}\dfrac43\).

  4. Exercise 14

    Prove that sin⁡−1817+sin⁡−135=sin⁡−17785\displaystyle \sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{77}{85}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A=\sin^{-1}\tfrac{8}{17},\qquad B=\sin^{-1}\tfrac35,\qquad A,B\in\left(0,\tfrac{\pi}{2}\right) \] \[\cos A=\tfrac{15}{17},\qquad \cos B=\tfrac45 \] \[\sin(A+B)=\sin A\cos B+\cos A\sin B=\tfrac{8}{17}\cdot\tfrac45+\tfrac{15}{17}\cdot\tfrac35=\tfrac{77}{85} \] \[\cos(A+B)=\cos A\cos B-\sin A\sin B=\tfrac{60}{85}-\tfrac{24}{85}=\tfrac{36}{85}>0 \] So \(\displaystyle A+B\in\left(0,\tfrac{\pi}{2}\right)\), where \(\displaystyle \sin^{-1}\) inverts \(\displaystyle \sin\). Answer: \(\displaystyle A+B=\sin^{-1}\dfrac{77}{85}\).

  5. Exercise 15

    Show that sin⁡−1513+cos⁡−135=tan⁡−16316\displaystyle \sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}=\tan ^{-1} \frac{63}{16}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[A=\sin^{-1}\tfrac{5}{13}\ \Rightarrow\ \cos A=\tfrac{12}{13},\quad \tan A=\tfrac{5}{12} \] \[B=\cos^{-1}\tfrac35\ \Rightarrow\ \sin B=\tfrac45,\quad \tan B=\tfrac43 \] \[\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\dfrac{5}{12}+\frac43}{1-\dfrac{5}{12}\cdot\frac43}=\frac{\dfrac{21}{12}}{\dfrac{16}{36}}=\frac{63}{16} \] Here \(\displaystyle 1-\tan A\tan B=\tfrac49>0\) with \(\displaystyle A,B\) acute, so \(\displaystyle A+B\in\left(0,\tfrac{\pi}{2}\right)\). Answer: \(\displaystyle A+B=\tan^{-1}\dfrac{63}{16}\).

  6. Exercise 16

    Prove that tan⁡−114+tan⁡−129=sin⁡−115\displaystyle \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\sin ^{-1} \frac{1}{\sqrt{5}}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\qquad (xy<1) \] \[\tan^{-1}\tfrac14+\tan^{-1}\tfrac29=\tan^{-1}\frac{\frac14+\frac29}{1-\frac14\cdot\frac29}=\tan^{-1}\frac{\dfrac{17}{36}}{\dfrac{17}{18}}=\tan^{-1}\tfrac12 \] Let \(\displaystyle \theta=\tan^{-1}\tfrac12\in\left(0,\tfrac{\pi}{2}\right)\): \[\sin\theta=\frac{\tan\theta}{\sqrt{1+\tan^{2}\theta}}=\frac{\frac12}{\dfrac{\sqrt5}{2}}=\frac{1}{\sqrt5} \] \[\theta=\sin^{-1}\frac{1}{\sqrt5} \] Answer: \(\displaystyle \tan^{-1}\dfrac14+\tan^{-1}\dfrac29=\tan^{-1}\dfrac12=\sin^{-1}\dfrac{1}{\sqrt5}\).

  7. Exercise 17

    Find the value of 4tan⁡−115−tan⁡−11239\displaystyle 4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{\pi}{4}\)
    Using \(\displaystyle 2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}\) for \(\displaystyle |x|<1\): \[2\tan^{-1}\tfrac15=\tan^{-1}\frac{2/5}{1-1/25}=\tan^{-1}\frac{5}{12} \] \[4\tan^{-1}\tfrac15=2\tan^{-1}\tfrac5{12}=\tan^{-1}\frac{5/6}{1-25/144}=\tan^{-1}\frac{120}{119} \] Using \(\displaystyle \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}\) for \(\displaystyle xy>-1\): \[\tan^{-1}\frac{120}{119}-\tan^{-1}\frac1{239}=\tan^{-1}\frac{\dfrac{120}{119}-\frac1{239}}{1+\dfrac{120}{119}\cdot\frac1{239}} \] \[=\tan^{-1}\frac{120\cdot239-119}{119\cdot239+120}=\tan^{-1}\frac{28561}{28561}=\tan^{-1}1 \] Answer: \(\displaystyle \dfrac{\pi}{4}\)

  8. Exercise 18

    Show that tan⁡(12sin⁡−134)=4−73\displaystyle \tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right)=\frac{4-\sqrt{7}}{3} and justify why the other value 4+73\displaystyle \frac{4+\sqrt{7}}{3} is ignored?

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle \theta=\tfrac12\sin^{-1}\tfrac34\) and \(\displaystyle t=\tan\theta\). Since \(\displaystyle 0<\tfrac34<1\), the principal value lies in \(\displaystyle \left(0,\tfrac\pi2\right)\), so \(\displaystyle 0<\theta<\tfrac\pi4\). \[\sin2\theta=\frac{2t}{1+t^2}=\frac34 \] \[3t^2-8t+3=0 \] \[t=\frac{8\pm\sqrt{64-36}}{6}=\frac{4\pm\sqrt7}{3} \] \[\frac{4+\sqrt7}{3}>\frac{4+2}{3}=2>1 \qquad 0<\frac{4-\sqrt7}{3}<1 \] For \(\displaystyle 0<\theta<\tfrac\pi4\) we need \(\displaystyle 0<\tan\theta<1\), so \(\displaystyle \frac{4+\sqrt7}{3}\) is rejected. Answer: \(\displaystyle \tan\left(\tfrac12\sin^{-1}\tfrac34\right)=\dfrac{4-\sqrt7}{3}\)

  9. Exercise 19

    If a1,a2,a3,…,an\displaystyle a_1, a_2, a_3, \ldots, a_n is an arithmetic progression with common difference d\displaystyle d, then evaluate the following expression. tan⁡[tan⁡−1(d1+a1a2)+tan⁡−1(d1+a2a3)+tan⁡−1(d1+a3a4)+…+tan⁡−1(d1+an−1an)].\tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)+\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right)+\ldots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_n}\right)\right] .

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{a_n-a_1}{1+a_1 a_n}\)
    The common difference gives \[d=a_{k+1}-a_k \] \[\tan^{-1}\frac{d}{1+a_ka_{k+1}}=\tan^{-1}\frac{a_{k+1}-a_k}{1+a_ka_{k+1}}=\tan^{-1}a_{k+1}-\tan^{-1}a_k \] Adding for \(\displaystyle k=1,2,\ldots,n-1\), the terms cancel in pairs: \[S=(\tan^{-1}a_2-\tan^{-1}a_1)+(\tan^{-1}a_3-\tan^{-1}a_2)+\cdots+(\tan^{-1}a_n-\tan^{-1}a_{n-1}) \] \[S=\tan^{-1}a_n-\tan^{-1}a_1 \] \[\tan S=\frac{a_n-a_1}{1+a_1a_n}=\frac{(n-1)d}{1+a_1a_n} \] Answer: \(\displaystyle \dfrac{(n-1)d}{1+a_1a_n}\)
  10. Choose the correct answers from the given four options in each of the Exercises from $\displaystyle 20$ to $\displaystyle 37$ (M.C.Q.).

    Exercise 20

    Which of the following is the principal value branch of cos⁡−1x\displaystyle \cos ^{-1} x ?
    (A)
    [−π2,π2]\displaystyle \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]
    (B)
    (0,π)\displaystyle (0, \pi)
    (C)
    [0,π]\displaystyle [0, \pi]
    (D)
    (0,π)−{π2}\displaystyle (0, \pi)-\left\{\frac{\pi}{2}\right\}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    C
    (C) \(\displaystyle [0,\pi]\) \[\cos^{-1}:[-1,1]\to[0,\pi] \] \(\displaystyle \cos x\) is one-one and onto \(\displaystyle [-1,1]\) on \(\displaystyle [0,\pi]\).