NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions - Exercise 2.1
Access Exercises of Class 12 Maths Chapter 2 – Inverse Trigonometric Functions
Exercise 2.1 Solutions: 14 Questions (12 Short Answers, 2 MCQs)
Exercise 2.2 Solutions: 21 Questions (18 Short Answers, 3 MCQs)
Miscellaneous Exercise Solutions: 17 Questions (14 Short Answers, 3 MCQs)
Exercise 2.1
Find the principal values of the following questions :
$$\textbf{1.}\space\textbf{sin}^{\normalsize-1}\bigg(-\frac{\textbf{1}}{\textbf{2}}\bigg)\\\textbf{2.\space}\textbf{cos}^{\normalsize-1}\bigg(\frac{\sqrt{\textbf{3}}}{\textbf{2}}\bigg)$$
3. cosec–1 (2)
$$\textbf{4. tan}^{\normalsize-1}(-\sqrt{\textbf{3}})\\\textbf{5.\space}\textbf{cos}^{\normalsize-1}\bigg(\textbf{-}\frac{\textbf{1}}{\textbf{2}}\bigg)$$
6. tan–1(– 1)
$$\textbf{7.\space}\textbf{sec}^{\normalsize-1}\bigg(\frac{\textbf{2}}{\sqrt{\textbf{3}}}\bigg)\\\textbf{8.\space}\textbf{cot}^{\normalsize-1}(\sqrt{\textbf{3}})\\\textbf{9.\space}\textbf{cos}^{-1}\bigg(-\frac{\textbf{1}}{\sqrt{\textbf{2}}}\bigg)\\\textbf{10.\space}\space \textbf{cosec}^{\normalsize-1}(-\sqrt{\textbf{2}})$$
Solutions :
$$\text{1.\space Let}\space\text{sin}^{-1}\bigg(-\frac{1}{2}\bigg)=\theta\\\Rarr\space\text{sin}\space\theta=-\frac{1}{2}$$
We know that the range of principal value of
$$\text{sin}^{\normalsize-1}\space\theta\space\text{is}\space\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].\\\therefore\space\text{sin}\space\theta =-\frac{1}{2}=-\text{sin}\frac{\pi}{6}\\=\text{sin}\bigg(-\frac{\pi}{6}\bigg) $$
[∵ sin (– θ) = – sin θ]
$$\theta = -\frac{\pi}{6},\text{where\space}\theta\epsilon\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]\\\Rarr\space\text{sin}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)=-\frac{\pi}{6}$$
Hence, the principal value of
$$\text{sin}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)\text{is}\space-\frac{\pi}{6}.$$
Note : Principal value of any inverse function is unique.
$$\text{2.\space Let cos}^{\normalsize-1}\bigg(\frac{\sqrt{3}}{2}\bigg)=\theta\\\Rarr\space\text{cos}\space\theta =\frac{\sqrt{3}}{2} $$
We know that the range of principal value of cos–1
θ is [0, π].
$$\therefore\space\text{cos}\space\theta = \frac{\sqrt{3}}{2}=\text{cos}\frac{3}{6}\\\Rarr\space\theta = \frac{\pi}{6},\text{where}\space\theta\epsilon[0,\pi]\\\Rarr\space\text{cos}^{\normalsize-1}\bigg(\frac{\sqrt{3}}{2}\bigg)=\frac{\pi}{6}$$
Hence, the principal value of
$$\text{cos}^{\normalsize-1}\bigg(\frac{\sqrt{3}}{2}\bigg)\text{is}\frac{\pi}{6}.$$
$$\text{3. Let cosec}^{\normalsize-1}\space2=\theta\\\Rarr\space \text{cosec}\space\theta=2.$$We know that the range of principal value of cosec–1 θ is
$$\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]-[0].$$
$$\text{cosec}\space\theta = 2 =\text{cosec}\frac{\pi}{6}\\\Rarr\space\theta =\frac{\pi}{6},\text{where}\space\theta\epsilon\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]-[0]\\\Rarr\space\text{cosec}^{\normalsize-1}(2)=\frac{\pi}{6}$$
Hence, the principal value of
$$\Rarr\text{cosec}^{\normalsize-1}(2)\text{is}\frac{\pi}{6}.$$
$$\text{4. Let tan}^{\normalsize-1}(-\sqrt{3})=\theta\\\Rarr\space\text{tan}\space\theta = -\sqrt{3}.$$
We know that the range of principal value of tan–1
$$\theta\space\text{is}\space\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)\\\therefore\space\text{tan}\space\theta=- \sqrt{3}=-\text{tan}\frac{\pi}{3}\\=\text{tan}\bigg(-\frac{\pi}{3}\bigg)\\\lbrack\because\space \text{tan}(-\theta) = -\text{tan}\space\theta\rbrack\\\Rarr\space\theta= -\frac{\pi}{3},\\\text{where}\space\theta\epsilon\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)\\\Rarr\space \text{tan}^{\normalsize-1}(-\sqrt{3})=-\frac{\pi}{3}$$
Hence, principal value of
$$\text{tan}^{\normalsize-1}(-\sqrt{3})\space\text{is}\space-\frac{\pi}{3}.$$
$$\text{5. Let cos}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)=\theta\\\Rarr\space\text{cos}\space\theta = -\frac{1}{2}$$
We know that the range of principal value of cos–1 θ is [0, π].
$$\therefore\space\text{cos}\space\theta=-\frac{1}{2}\\=-\text{cos}\frac{\pi}{3}=\text{cos}\bigg(\pi-\frac{\pi}{3}\bigg)\\\lbrack\because\space \text{cos}(\pi-\theta)=-\text{cos}\space\theta\rbrack\\=\text{cos}\frac{2\pi}{3}\\\Rarr\space\theta =\frac{2\pi}{3},\text{where}\theta\epsilon[0,\pi]\\\Rarr\text{cos}^{\normalsize-1} =\bigg(-\frac{1}{2}\bigg)=\frac{2\pi}{3}$$
Hence, principal value of
$$\text{cos}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)\text{is}\frac{2\pi}{3}.$$
Note : cos–1(– θ) ≠ – cos–1 θ
6. Let tan–1(– 1) = θ
$$\Rarr\space \text{tan}\space\theta=-1$$
We know that the range of principal value of tan–1 θ is
$$\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg).\\\therefore\space\text{tan}\space\theta=-1=-\text{tan}\frac{\pi}{4}\\=\text{tan}\bigg(-\frac{\pi}{4}\bigg)\\\lbrack\because\space\text{tan}(-\theta) =-\text{tan}\space\theta\rbrack\\\Rarr\space\theta = \frac{\pi}{4},\text{where}\space\theta\epsilon\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)\\\bigg[\therefore\space\text{tan}^{\normalsize-1}(\normalsize-1)=-\frac{\pi}{4}\bigg]$$
Hence, the principal value of tan–1 (– 1) is
$$-\frac{\pi}{4}.$$
$$\text{7. Let sec}^{\normalsize-1}\bigg(\frac{2}{\sqrt{3}}\bigg)=\theta\\\Rarr\space\text{sec}\space\theta=\frac{2}{\sqrt{3}}$$
We know that the range of principal value of sec–1 θ is
$$[0,\pi]-\lbrace\frac{\pi}{2}\rbrace.\\\therefore\space\text{sec}\space\theta=\frac{2}{\sqrt{3}}=\text{sec}\frac{\pi}{6}\\\Rarr\space \theta=\frac{\pi}{6},\\\text{where}\space\theta\epsilon[0,\pi]-\begin{Bmatrix}\frac{\pi}{2}\end{Bmatrix}\\\Rarr\space\text{sec}^{\normalsize-1}\bigg(\frac{2}{\sqrt{3}}\bigg)=\frac{\pi}{6}\\\text{Hence, the principal value of}\\\text{sec}^{\normalsize-1}\bigg(\frac{2}{\sqrt{3}}\bigg)\text{is}\frac{\pi}{6}.$$
$$\text{8. Let cot}^{\normalsize-1}(\sqrt{3})=\theta\\\Rarr\space\text{cot}\space\theta=\sqrt{3}$$
We know that the range of principal value of cot–1 θ is (0, π).
$$\therefore\space\text{cot}\space\theta=\sqrt{3}=\text{cot}\frac{\pi}{6}\\\Rarr\space\theta=\frac{\pi}{6},\space\text{where}\pi\space\epsilon(0,\pi)\\\Rarr\space\text{cot}^{\normalsize-1}(\sqrt{3})=\frac{\pi}{6}\\\text{Hence, the principal value of}\\\text{cot}^{\normalsize-1}(\sqrt{3})\space\text{is}\space\frac{\pi}{6}.$$
$$\text{9. Let}\space\text{cos}^{\normalsize-1}\bigg(-\frac{1}{\sqrt{2}}\bigg)=\theta\\\Rarr\space\text{cos}\space\theta = -\frac{1}{\sqrt{2}}$$
We know that the range of principal value of cos–1 θ is [0, π].
$$\therefore\space\text{cos}\space\theta =-\frac{1}{\sqrt{2}}\\=-\text{cos}\frac{\pi}{4}=\text{cos}\bigg(\pi-\frac{3}{4}\bigg)\\\lbrack\because\space\text{cos}(\pi-\theta)=-\theta\rbrack\\\Rarr\space\theta =\frac{3\pi}{4},\text{where}\space\theta\epsilon[0,\pi]\\\Rarr\space\text{cos}^{\normalsize-1}\bigg(-\frac{1}{\sqrt{2}}\bigg)\text{is}\space\frac{3\pi}{4}$$
Hence, the principal value of
$$\text{cos}^{\normalsize-1}\bigg(-\frac{1}{\sqrt{2}}\bigg)\text{is}\frac{3\pi}{4}.$$
$$\text{10.\space Let}\space\text{cosec}^{\normalsize-1}(-\sqrt{2})=\theta$$
We know that the range of principal value of
$$\text{cosec}^{\normalsize-1}\theta\space\text{is}\space\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]-[0].\\\therefore\space\text{cosec}\space\theta =-\sqrt{2}\\=\text{cosec}\frac{\pi}{4} = \text{cosec}\bigg(-\frac{\pi}{4}\bigg)\\\lbrack\because\space\text{cosec}(-\theta) =-\text{cosec}\space\theta\rbrack\\\Rarr\theta = -\frac{\pi}{4},\\\text{where}\space\theta\epsilon\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]-[0]\\\Rarr\space \text{cosec}^{\normalsize-1}(-\sqrt{2})=-\frac{\pi}{4}\\\text{Hence, the principal value of}\\\text{cosec}^{\normalsize-1}(-\sqrt{2})=-\frac{\pi}{4}.$$
Direction (Q. 11 to 14) : Find the values of the following questions :
$$\textbf{11.\space tan}^{\normalsize-1}\textbf{(1) + cos}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg) \textbf{+}\\\textbf{sin}^{\normalsize-1}\bigg(-\frac{\textbf{1}}{\textbf{2}}\bigg)\textbf{.}$$
Given expression is not a standard identity, so we separately find the value of
$$\textbf{tan}^{\textbf{-1}}(\textbf{1})\textbf{,}\space\textbf{cos}^{\normalsize-1}\bigg(-\frac{\textbf{1}}{\textbf{2}}\bigg)\textbf{,}\\\textbf{sin}^{\normalsize-1}\bigg(\textbf{-}\frac{\textbf{1}}{\textbf{2}}\bigg)\\\textbf{and simplify it.}$$
Sol. Let tan–1(1) = x
$$\Rarr\space\text{tan x}=1=\text{tan}\frac{\pi}{4}\\\Rarr\space x=\frac{\pi}{4}\\\text{where principal value}\\x\epsilon\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)\\\therefore\space\text{tan}^{\normalsize-1}(1)=\frac{\pi}{4}\\\text{Let cos}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)=y\\\Rarr\space\text{cos y}=-\frac{1}{2}=-\text{cos}\bigg(\frac{\pi}{3}\bigg)\\=\text{cos}\bigg(\pi-\frac{\pi}{3}\bigg)=\text{cos}\bigg(\frac{2\pi}{3}\bigg)$$
[∵ cos (π – θ) = – cos θ]
$$\Rarr\space y =\frac{2\pi}{3},\\\space\text{where principal value y ∈ [0, π].}\\\therefore\space\text{cos}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)=\frac{2\pi}{3}\\\text{Let\space}\text{sin}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)=z\\\Rarr\space\text{sin z} =\frac{1}{2}=-\text{sin}\bigg(\frac{\pi}{6}\bigg)\\\text{sin}\bigg(-\frac{\pi}{6}\bigg)\\\Rarr\space z = -\frac{\pi}{6},\text{where principal value}\\z\epsilon\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$$
$$\therefore\space\text{sin}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg) = -\frac{\pi}{6}\\\therefore\space\text{tan}^{-1}(1) + \text{cos}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg) +\\\text{sin}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)\\= x +y+z=\\\frac{\pi}{4} + \frac{2\pi}{3}- \frac{\pi}{6}\\=\frac{\pi}{4} + \frac{2\pi}{3}-\frac{\pi}{6}\\=\frac{3\pi+ 8\pi-2\pi}{12}=\frac{9\pi}{12}\\=\frac{3\pi}{4}$$
$$\textbf{12.}\space\textbf{cos}^{\normalsize-1}\bigg(\frac{\textbf{1}}{\textbf{2}}\bigg) \textbf{+ 2}\textbf{sin}^{\textbf{-1}}\bigg(\frac{\textbf{1}}{\textbf{2}}\bigg)\textbf{.}$$
Sol. We can find the value of given expression by simplifying the individual terms.
$$\text{Let\space}\text{cos}^{\normalsize-1}\bigg(\frac{1}{2}\bigg)=x\\\Rarr\space \text{cos x}=\frac{1}{2}=\text{cos}\frac{\pi}{3}\\\Rarr\space x=\frac{\pi}{3}\epsilon[0,\pi]$$
(principal interval)
Again, let
$$\text{sin}^{\normalsize-1}\bigg(\frac{1}{2}\bigg)=y\\\Rarr\space\text{sin y}= \frac{1}{2}=\text{sin}\frac{\pi}{6}\\\Rarr\space y =\frac{\pi}{6}\epsilon\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$$
(principal interval)
$$\therefore\space\text{cos}^{\normalsize-1}\bigg(\frac{1}{2}\bigg) + 2\text{sin}^{\normalsize-1}\bigg(\frac{1}{2}\bigg)\\= x + 2y\\=\frac{\pi}{3}+2×\frac{\pi}{6} =\frac{2\pi}{3}$$
13. If sin–1 x = y, then
$$\textbf{(a)\space}\textbf{0}\leq \textbf{y}\leq\pi\\\textbf{(b)\space}-\frac{\pi}{\textbf{2}}\leq \textbf{y}\leq\frac{\pi}{\textbf{2}}\\\textbf{(c)\space 0}\lt \textbf{y}\lt\pi\\\textbf{(d)\space}-\frac{\pi}{\textbf{2}}\lt \textbf{y}\lt\frac{\pi}{\textbf{2}}\\\textbf{Sol.\space}(b)\space-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\\\text{As range of sin}^{\normalsize–1}x is\space\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg],\\\text{therefore}-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}.$$
$$\textbf{14.\space tan}^{\normalsize-1}\sqrt{\textbf{3}}-\textbf{sec}^{\normalsize-1}(\normalsize-\textbf{2})\space\\\textbf{is equal to :}\\\textbf{(a)\space }\pi\\\textbf{(b)\space}-\frac{\pi}{\textbf{3}}\\\textbf{(c)\space}\frac{\pi}{\textbf{3}}\\\textbf{(d)\space}\frac{\textbf{2}\pi}{\textbf{3}}\\\textbf{Sol.\space}(b)\space-\frac{\pi}{3}\\\text{Let}\space\text{tan}^{\normalsize-1}\sqrt{3}=x\\\Rarr\space\text{tan x =}\sqrt{3}\\\Rarr\space \text{tan x = tan}\frac{\pi}{3}$$
$$\Rarr\space x=\frac{\pi}{3}\epsilon\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg)\\\text{(principal interval)}$$
Let sec–1(– 2) = y
$$\Rarr\space \text{sec y}=-2\\\Rarr\space\text{sec y= -sec}\frac{\pi}{3}\\\Rarr\space\text{sec y= sec}\bigg(\pi-\frac{\pi}{3}\bigg)\\\lbrack\because\space\text{sec}(\pi-\theta) =-\text{sec}\space\theta\rbrack\\\Rarr\space\text{ sec y = sec}\bigg(\frac{2\pi}{3}\bigg)\\ y = \frac{2\pi}{3}\epsilon[0,\pi]\\\text{(principal interval)}\\\therefore\space\text{tan}^{\normalsize-1}\sqrt{3}-\text{sec}^{\normalsize-1}(-2)$$
= x - y
$$=\frac{\pi}{3}-\frac{2\pi}{3}\\=-\frac{\pi}{3}$$
