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Inverse Trigonometric Functions

CBSE · Class 12 · Mathematics

NCERT Solutions for Inverse Trigonometric Functions — CBSE Class 12 Mathematics.

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42 Questions Solved · 3 Sections

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EXERCISE 2.1

1sin⁡−1(−12)\sin^{-1}\left(-\frac{1}{2}\right)Show solution
The principal value range of **sin⁡−1x\sin^{-1}x** is [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. Since
sin⁡(−π6)=−12, \sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2},
we get
sin⁡−1(−12)=−π6. \sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}.

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2cos⁡−1(32)\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)Show solution
The principal value range of **cos⁡−1x\cos^{-1}x** is [0,π][0,\pi]. Since
cos⁡(π6)=32, \cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2},
therefore
cos⁡−1(32)=π6. \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}.

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3cosec−1(2)\text{cosec}^{-1}(2)Show solution
For **cosec⁡−1x\cosec^{-1}x**, the principal value range is [−π2,π2] ⁣−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\! - \{0\}. We need cosec⁡y=2\cosec y=2, so
sin⁡y=12. \sin y=\frac{1}{2}.
In the principal range, this gives
y=π6. y=\frac{\pi}{6}.
Hence,
cosec⁡−1(2)=π6. \cosec^{-1}(2)=\frac{\pi}{6}.

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4tan⁡−1(−3)\tan^{-1}(-\sqrt{3})Show solution
The principal value range of **tan⁡−1x\tan^{-1}x** is (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right). Since
tan⁡(−π3)=−3, \tan\left(-\frac{\pi}{3}\right)=-\sqrt{3},
we have
tan⁡−1(−3)=−π3. \tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}.

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5cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{2}\right)Show solution
The principal value range of **cos⁡−1x\cos^{-1}x** is [0,π][0,\pi]. Since
cos⁡(2π3)=−12, \cos\left(\frac{2\pi}{3}\right)=-\frac{1}{2},
therefore
cos⁡−1(−12)=2π3. \cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}.

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6tan⁡−1(−1)\tan^{-1}(-1)Show solution
The principal value range of **tan⁡−1x\tan^{-1}x** is (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right). Since
tan⁡(−π4)=−1, \tan\left(-\frac{\pi}{4}\right)=-1,
we get
tan⁡−1(−1)=−π4. \tan^{-1}(-1)=-\frac{\pi}{4}.

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7sec⁡−1(23)\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)Show solution
For **sec⁡−1x\sec^{-1}x**, the principal value range is [0,π]−{π2}[0,\pi]-\left\{\frac{\pi}{2}\right\}. We need
sec⁡y=23⇒cos⁡y=32. \sec y=\frac{2}{\sqrt{3}} \quad\Rightarrow\quad \cos y=\frac{\sqrt{3}}{2}.
In the principal range, this gives
y=π6. y=\frac{\pi}{6}.
So,
sec⁡−1(23)=π6. \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}.

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8cot⁡−1(3)\cot^{-1}(\sqrt{3})Show solution
The principal value range of **cot⁡−1x\cot^{-1}x** is (0,π)(0,\pi). Since
cot⁡(π6)=3, \cot\left(\frac{\pi}{6}\right)=\sqrt{3},
therefore
cot⁡−1(3)=π6. \cot^{-1}(\sqrt{3})=\frac{\pi}{6}.

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9cos⁡−1(−12)\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)Show solution
The principal value range of **cos⁡−1x\cos^{-1}x** is [0,π][0,\pi]. Since
cos⁡(3π4)=−12, \cos\left(\frac{3\pi}{4}\right)=-\frac{1}{\sqrt{2}},
we get
cos⁡−1(−12)=3π4. \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi}{4}.

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10cosec⁡−1(−2)\operatorname{cosec}^{-1}(-\sqrt{2})Show solution
For **cosec⁡−1x\cosec^{-1}x**, the principal value range is [−π2,π2]−{0}\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}. We need
cosec⁡y=−2⇒sin⁡y=−12. \cosec y=-\sqrt{2} \quad\Rightarrow\quad \sin y=-\frac{1}{\sqrt{2}}.
In the principal range, this gives
y=−3π4 y=-\frac{3\pi}{4}
since sin⁡(−3π4)=−12\sin\left(-\frac{3\pi}{4}\right)=-\frac{1}{\sqrt{2}}. Hence,
cosec⁡−1(−2)=−3π4. \cosec^{-1}(-\sqrt{2})=-\frac{3\pi}{4}.

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11tan⁡−1(1)+cos⁡−1−12+sin⁡−1−12\tan^{-1}(1) + \cos^{-1} - \frac{1}{2} + \sin^{-1} - \frac{1}{2}Show solution
Using principal values:
tan⁡−1(1)=π4,cos⁡−1(−12)=2π3,sin⁡−1(−12)=−π6. \tan^{-1}(1)=\frac{\pi}{4},\quad \cos^{-1}\left(-\frac12\right)=\frac{2\pi}{3},\quad \sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}.
So the value is
π4+2π3−π6=π4+4π6−π6=π4+3π6=π4+π2=3π4. \frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6} =\frac{\pi}{4}+\frac{4\pi}{6}-\frac{\pi}{6} =\frac{\pi}{4}+\frac{3\pi}{6} =\frac{\pi}{4}+\frac{\pi}{2} =\frac{3\pi}{4}.
However, this expression is printed in the book as a value-finding exercise with the intended grouping **tan⁡−1(1)+cos⁡−1(−12)+sin⁡−1(−12)\tan^{-1}(1)+\cos^{-1}(-\tfrac12)+\sin^{-1}(-\tfrac12)**, which gives
π4+2π3−π6=3π4. \frac{\pi}{4}+\frac{2\pi}{3}-\frac{\pi}{6}=\frac{3\pi}{4}.
So the computed value is 3π4\frac{3\pi}{4}. If the expression was meant differently, the book text is ambiguous. Based on the printed chapter notation, the sum evaluates to 3π4\frac{3\pi}{4}.

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12cos⁡−112+2sin⁡−112\cos^{-1}\frac{1}{2} + 2\sin^{-1}\frac{1}{2}Show solution

cos⁡−1(12)=π3,sin⁡−1(12)=π6. \cos^{-1}\left(\frac12\right)=\frac{\pi}{3},\qquad \sin^{-1}\left(\frac12\right)=\frac{\pi}{6}.
Therefore,
cos⁡−1(12)+2sin⁡−1(12)=π3+2⋅π6=π3+π3=2π3. \cos^{-1}\left(\frac12\right)+2\sin^{-1}\left(\frac12\right) =\frac{\pi}{3}+2\cdot\frac{\pi}{6} =\frac{\pi}{3}+\frac{\pi}{3} =\frac{2\pi}{3}.
But the printed chapter line shows the expression as cos⁡−112+2sin⁡−112\cos^{-1}\frac{1}{2}+2\sin^{-1}\frac{1}{2}, which equals 2π3\frac{2\pi}{3}.

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13If sin⁡−1x=y\sin^{-1}x = y, thenShow solution
From the chapter, the principal value branch of **sin⁡−1x\sin^{-1}x** has range
[−π2,π2]. \left[-\frac{\pi}{2},\frac{\pi}{2}\right].
So if sin⁡−1x=y\sin^{-1}x=y, then
−π2≤y≤π2. -\frac{\pi}{2}\le y\le \frac{\pi}{2}.
Hence option (B) is correct.

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14tan⁡−13−sec⁡−1(−2)\tan^{-1}\sqrt{3} - \sec^{-1}(-2) is equal toShow solution
We use principal values:
tan⁡−1(3)=π3. \tan^{-1}(\sqrt{3})=\frac{\pi}{3}.
Also, for **sec⁡−1(−2)\sec^{-1}(-2)**, we need sec⁡y=−2\sec y=-2, i.e.
cos⁡y=−12. \cos y=-\frac12.
In the principal range of **sec⁡−1\sec^{-1}**, which is [0,π]−{π2}[0,\pi]-\left\{\frac{\pi}{2}\right\}, this gives
y=2π3. y=\frac{2\pi}{3}.
Therefore,
tan⁡−1(3)−sec⁡−1(−2)=π3−2π3=−π3. \tan^{-1}(\sqrt{3})-\sec^{-1}(-2)=\frac{\pi}{3}-\frac{2\pi}{3}=-\frac{\pi}{3}.
So the correct option is (B), and the value is −π3-\frac{\pi}{3}.

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EXERCISE 2.2

13sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3\sin^{-1}x = \sin^{-1}(3x - 4x^3), x \in \left[-\frac{1}{2}, \frac{1}{2}\right]Show solution
Let x=sin⁡θx=\sin\theta, where θ=sin⁡−1x\theta=\sin^{-1}x. Then
2x1−x2=2sin⁡θcos⁡θ=sin⁡2θ. 2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin 2\theta.
So
sin⁡−1(2x1−x2)=sin⁡−1(sin⁡2θ). \sin^{-1}(2x\sqrt{1-x^2})=\sin^{-1}(\sin 2\theta).
For x∈[−12,12]x\in\left[-\frac12,\frac12\right], we have θ∈[−π6,π6]\theta\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right], hence 2θ∈[−π3,π3]2\theta\in\left[-\frac{\pi}{3},\frac{\pi}{3}\right], which lies in the principal range of sin⁡−1\sin^{-1}. Therefore
sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1x. \sin^{-1}(\sin 2\theta)=2\theta=2\sin^{-1}x.
Hence proved.

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23cos⁡−1x=cos⁡−1(4x3−3x),x∈[12,1]3\cos^{-1}x = \cos^{-1}(4x^3 - 3x), x \in \left[\frac{1}{2}, 1\right]Show solution
Let x=cos⁡θx=\cos\theta, where θ=cos⁡−1x\theta=\cos^{-1}x. Then
4x3−3x=4cos⁡3θ−3cos⁡θ=cos⁡3θ. 4x^3-3x=4\cos^3\theta-3\cos\theta=\cos 3\theta.
So
cos⁡−1(4x3−3x)=cos⁡−1(cos⁡3θ). \cos^{-1}(4x^3-3x)=\cos^{-1}(\cos 3\theta).
For x∈[12,1]x\in\left[\frac12,1\right], we have θ∈[0,π3]\theta\in[0,\frac{\pi}{3}], hence 3θ∈[0,π]3\theta\in[0,\pi], the principal range of cos⁡−1\cos^{-1}. Therefore
cos⁡−1(cos⁡3θ)=3θ=3cos⁡−1x. \cos^{-1}(\cos 3\theta)=3\theta=3\cos^{-1}x.
Hence proved.

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3tan⁡−11+x2−1x,x≠0\tan^{-1}\frac{\sqrt{1 + x^2} - 1}{x}, x \neq 0Show solution
Let
tan⁡θ=x⇒θ=tan⁡−1x. \tan\theta=x \quad\Rightarrow\quad \theta=\tan^{-1}x.
Then
1+x2=1+tan⁡2θ=sec⁡θ. \sqrt{1+x^2}=\sqrt{1+\tan^2\theta}=\sec\theta.
So
1+x2−1x=sec⁡θ−1tan⁡θ. \frac{\sqrt{1+x^2}-1}{x}=\frac{\sec\theta-1}{\tan\theta}.
Multiplying numerator and denominator by sec⁡θ+1\sec\theta+1,
sec⁡θ−1tan⁡θ=sec⁡2θ−1tan⁡θ(sec⁡θ+1)=tan⁡2θtan⁡θ(sec⁡θ+1)=tan⁡θsec⁡θ+1. \frac{\sec\theta-1}{\tan\theta}=\frac{\sec^2\theta-1}{\tan\theta(\sec\theta+1)}=\frac{\tan^2\theta}{\tan\theta(\sec\theta+1)}=\frac{\tan\theta}{\sec\theta+1}.
Using tan⁡θ=sin⁡θcos⁡θ\tan\theta=\frac{\sin\theta}{\cos\theta} and sec⁡θ+1=1+cos⁡θcos⁡θ\sec\theta+1=\frac{1+\cos\theta}{\cos\theta},
tan⁡θsec⁡θ+1=sin⁡θ1+cos⁡θ=tan⁡θ2. \frac{\tan\theta}{\sec\theta+1}=\frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2}.
Hence
tan⁡−1(1+x2−1x)=tan⁡−1(tan⁡θ2)=θ2=12tan⁡−1x. \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\tan^{-1}(\tan \tfrac{\theta}{2})=\frac{\theta}{2}=\frac12\tan^{-1}x.

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4tan⁡−1(1−cos⁡x1+cos⁡x),0<x<π\tan^{-1}\left(\sqrt{\frac{1 - \cos x}{1 + \cos x}}\right), 0 < x < \piShow solution
Use the identity
tan⁡x2=1−cos⁡x1+cos⁡x,0<x<π. \tan\frac{x}{2}=\sqrt{\frac{1-\cos x}{1+\cos x}}, \qquad 0<x<\pi.
Therefore,
tan⁡−1(1−cos⁡x1+cos⁡x)=tan⁡−1(tan⁡x2). \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right) =\tan^{-1}\left(\tan\frac{x}{2}\right).
Since 0<x<π0<x<\pi, we have 0<x2<π20<\frac{x}{2}<\frac{\pi}{2}, which is within the principal range of tan⁡−1\tan^{-1}. Hence
tan⁡−1(tan⁡x2)=x2. \tan^{-1}\left(\tan\frac{x}{2}\right)=\frac{x}{2}.

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5tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x),−π4<x<3π4\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right), \frac{-\pi}{4} < x < \frac{3\pi}{4}Show solution
Let t=tan⁡xt=\tan x. Then
cos⁡x−sin⁡xcos⁡x+sin⁡x=1−tan⁡x1+tan⁡x=tan⁡(π4−x). \frac{\cos x-\sin x}{\cos x+\sin x} =\frac{1-\tan x}{1+\tan x} =\tan\left(\frac{\pi}{4}-x\right).
Hence
tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)=tan⁡−1(tan⁡(π4−x)). \tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right) =\tan^{-1}\left(\tan\left(\frac{\pi}{4}-x\right)\right).
For −π4<x<3π4-\frac{\pi}{4}<x<\frac{3\pi}{4}, the angle π4−x\frac{\pi}{4}-x lies in (−π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right), so it is already in the principal range of tan⁡−1\tan^{-1}. Therefore
tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)=π4−x. \tan^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi}{4}-x.

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6tan⁡−1xa2−x2,∣x∣<a\tan^{-1}\frac{x}{\sqrt{a^2 - x^2}}, |x| < aShow solution
Let
x=asin⁡θwith −π2<θ<π2. x=a\sin\theta \quad\text{with } -\frac{\pi}{2}<\theta<\frac{\pi}{2}.
Then
a2−x2=a2−a2sin⁡2θ=acos⁡θ, \sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2\theta}=a\cos\theta,
so
xa2−x2=asin⁡θacos⁡θ=tan⁡θ. \frac{x}{\sqrt{a^2-x^2}}=\frac{a\sin\theta}{a\cos\theta}=\tan\theta.
Therefore
tan⁡−1(xa2−x2)=tan⁡−1(tan⁡θ)=θ. \tan^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right)=\tan^{-1}(\tan\theta)=\theta.
Since θ=sin⁡−1(xa)\theta=\sin^{-1}\left(\frac{x}{a}\right), the simplest form is
sin⁡−1(xa). \sin^{-1}\left(\frac{x}{a}\right).

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7tan⁡−1(3a2x−x3a3−3ax2),a>0;−a3<x<a3\tan^{-1}\left(\frac{3a^2x - x^3}{a^3 - 3ax^2}\right), a > 0; \frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}Show solution
Let
x=acos⁡θ. x=a\cos\theta.
Then
a3−3ax2 and 3a2x−x3 a^3-3ax^2 \text{ and } 3a^2x-x^3
are the standard forms appearing in the triple-angle identity
tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ. \tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}.
Taking
tan⁡θ=xa2−x2 \tan\theta=\frac{x}{\sqrt{a^2-x^2}}
or equivalently setting x=acos⁡θx=a\cos\theta, we get
3a2x−x3a3−3ax2=tan⁡3θ. \frac{3a^2x-x^3}{a^3-3ax^2}=\tan 3\theta.
Hence
tan⁡−1(3a2x−x3a3−3ax2)=3θ. \tan^{-1}\left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)=3\theta.
With θ=cos⁡−1(xa)\theta=\cos^{-1}\left(\frac{x}{a}\right) and the given interval, the principal-value result is
3cos⁡−1(xa). 3\cos^{-1}\left(\frac{x}{a}\right).

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8tan⁡−1[2cos⁡(2sin⁡−112)]\tan^{-1}\left[2\cos \left(2\sin^{-1}\frac{1}{2}\right)\right]
9tan⁡12[sin⁡−12x1+x2+cos⁡−11−y21+y2],∣x∣<1,y>0\tan \frac{1}{2}\left[\sin^{-1}\frac{2x}{1 + x^2} +\cos^{-1}\frac{1 - y^2}{1 + y^2}\right],|x| < 1,y > 0 and xy<1xy < 1
10sin⁡−1(sin⁡2π3)\sin^{-1}\left(\sin\frac{2\pi}{3}\right)
11tan⁡−1(tan⁡3π4)\tan^{-1}\left(\tan\frac{3\pi}{4}\right)
12tan⁡(sin⁡−135+cot⁡−132)\tan\left(\sin^{-1}\frac{3}{5}+\cot^{-1}\frac{3}{2}\right)
13cos⁡−1(cos⁡7π6)\cos^{-1}\left(\cos\frac{7\pi}{6}\right) is equal to
14sin⁡(π3−sin⁡−1(−12))\sin\left(\frac{\pi}{3}-\sin^{-1}\left(-\frac{1}{2}\right)\right) is equal to
15tan⁡−13−cot⁡−1(−3)\tan^{-1}\sqrt{3}-\cot^{-1}\left(-\sqrt{3}\right) is equal to

Miscellaneous Exercise on Chapter 2

1cos⁡−1(cos⁡13π6)\cos^{-1}\left(\cos\frac{13\pi}{6}\right)
2tan⁡−1(tan⁡7π6)\tan^{-1}\left(\tan\frac{7\pi}{6}\right)
32sin⁡−135=tan⁡−12472\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}
4sin⁡−1817+sin⁡−135=tan⁡−17736\sin^{-1}\frac{8}{17} + \sin^{-1}\frac{3}{5} = \tan^{-1}\frac{77}{36}
5cos⁡−145+cos⁡−11213=cos⁡−13365\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}
6cos⁡−11213+sin⁡−135=sin⁡−15665\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}
7tan⁡−16316=sin⁡−1513+cos⁡−135\tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5}
8tan⁡−1x=12cos⁡−11−x1+x,x∈[0,1]\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x}, x \in [0, 1]
9cot⁡−1(1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x)=x2,x∈(0,π4)\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}, x \in \left(0, \frac{\pi}{4}\right)
10tan⁡−1(1+x−1−x1+x+1−x)=π4−12cos⁡−1x,−12≤x≤1\tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x, -\frac{1}{\sqrt{2}} \leq x \leq 1 [Hint: Put x=cos⁡2θx = \cos 2\theta]
112tan⁡−1(cos⁡x)=tan⁡−1(2cosec⁡x)2\tan^{-1}(\cos x) = \tan^{-1}(2\operatorname{cosec} x) 12. tan⁡−11−x1+x=12tan⁡−1x,(x>0)\tan^{-1}\frac{1-x}{1+x} = \frac{1}{2}\tan^{-1}x, (x > 0)
13sin⁡(tan⁡−1x),∣x∣<1\sin(\tan^{-1}x), |x| < 1 is equal to
14sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x) - 2\sin^{-1}x = \frac{\pi}{2}, then xx is equal to

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