Exercise-2.2, Class 12th, Maths, Chapter -2, NCERT

EXERCISE 2.2 — Solutions

1. Prove:

$$3{\sin }^{-1}x={\sin }^{-1}\text{\hspace{0.17em}⁣}(3x-4x^3),x\in [-\frac{1}{2},\frac{1}{2}]\mathrm{}$$

Solution. Put $x=\sin \theta$ with $\theta ={\sin }^{-1}x$. For $x\in [-\frac{1}{2},\frac{1}{2}]$ we have $\theta \in [-\frac{\pi }{6},\frac{\pi }{6}]$, so $3\theta$ lies in the principal branch $[-\frac{\pi }{2},\frac{\pi }{2}]$. Using the triple-angle identity $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$ we get

$${\sin }^{-1}\text{\hspace{0.17em}⁣}(3x-4x^3)={\sin }^{-1}\text{\hspace{0.17em}⁣}(\sin 3\theta )=3\theta =3{\sin }^{-1}x,$$

as required.

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2. Prove:

$$3{\cos }^{-1}x={\cos }^{-1}\text{\hspace{0.17em}⁣}(4x^3-3x),x\in [\frac{1}{2},1]\mathrm{}$$

Solution. Put $x=\cos \varphi$ with $\varphi ={\cos }^{-1}x$. For $x\in [\frac{1}{2},1]$ we have $\varphi \in [0,\frac{\pi }{3}]$, so $3\varphi \in [0,\pi ]$ (principal branch for ${\cos }^{-1}$). Use $\cos 3\varphi =4{\cos }^3\varphi -3\cos \varphi$. Then

$${\cos }^{-1}\text{\hspace{0.17em}⁣}(4x^3-3x)={\cos }^{-1}\text{\hspace{0.17em}⁣}(\cos 3\varphi )=3\varphi =3{\cos }^{-1}x\mathrm{}$$

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3. Write in simplest form:

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{\sqrt{1+x^2}-1}{x}\right),x\mathrm{\ne }0$$

Solution. Put $x=\tan \theta$ so $\theta ={\tan }^{-1}x$ and $\sqrt{1+x^2}=\sec \theta$. Then

$$\frac{\sqrt{1+x^2}-1}{x}=\frac{\sec \theta -1}{\tan \theta }=\frac{(1-\cos \theta )\mathrm{/}\cos \theta }{\sin \theta \mathrm{/}\cos \theta }=\frac{1-\cos \theta }{\sin \theta }\mathrm{}$$

Now $1-\cos \theta =2{\sin }^2(\theta \mathrm{/}2)$ and $\sin \theta =2\sin (\theta \mathrm{/}2)\cos (\theta \mathrm{/}2)$, hence the ratio equals $\tan (\theta \mathrm{/}2)$. Therefore

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\theta \mathrm{/}2=\frac{1}{2}{\tan }^{-1}x\mathrm{.}$$

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4. Write in simplest form:

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),0<x<\pi \mathrm{}$$

Solution. Use the half-angle identity:

$$\frac{1-\cos x}{1+\cos x}={\tan }^2\frac{x}{2}\mathrm{}$$

Since $0<x<\pi$, $\tan (\frac{x}{2})>0$, so the square root equals $\tan (\frac{x}{2})$. Hence the arctangent is

$${\tan }^{-1}\text{\hspace{0.17em}⁣}(\tan \frac{x}{2})=\frac{x}{2}\mathrm{}$$

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5. Write in simplest form:

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right),-\frac{\pi }{4}<x<\frac{3\pi }{4}\mathrm{.}$$

Solution. Divide numerator and denominator by $\cos x$:

$$\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}=\tan (\frac{\pi }{4}-x)\mathrm{}$$

Principal-value constraints given ensure no ambiguity, so

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)=\frac{\pi }{4}-x\mathrm{}$$

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6. Write in simplest form:

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{x}{\sqrt{a^2-x^2}}\right),\mathrm{\mid }x\mathrm{\mid }<a\mathrm{}$$

Solution. Put $x=a\sin \theta$ (possible since $\mathrm{\mid }x\mathrm{\mid }<a$). Then $\sqrt{a^2-x^2}=a\cos \theta$ and the ratio is $\tan \theta$. Thus the expression equals

$${\tan }^{-1}(\tan \theta )=\theta ={\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{x}{a}\right)\mathrm{}$$

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7. Write in simplest form:

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right),a>0,-\frac{a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}\mathrm{}$$

Solution. Put $t={\displaystyle \frac{x}{a}}$. Then the argument becomes

$$\frac{3t-t^3}{1-3t^2}=\tan (3\alpha )\text{if }t=\tan \alpha \mathrm{.}$$

So put $x=a\tan \alpha$ (allowed in the given range). Then the inner expression is $\tan 3\alpha$ and the arctangent yields $3\alpha$. But $\alpha ={\tan }^{-1}(x\mathrm{/}a)$. Hence

$${\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{3a^{2}x-x^3}{a^3-3a{x}^2}\right)=3{\tan }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{x}{a}\right)\mathrm{}$$

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8. Find the value of:

$${\tan }^{-1}\text{\hspace{0.17em}⁣}(2\cos (2{\sin }^{-1}\frac{1}{2}))\mathrm{}$$

Solution. ${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$. So $2{\sin }^{-1}\frac{1}{2}=\frac{\pi }{3}$. Then $\cos (\frac{\pi }{3})=\frac{1}{2}$, hence the argument becomes $2\cdot \frac{1}{2}=1$. Therefore

$${\tan }^{-1}(1)=\frac{\pi }{4}\mathrm{}$$

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9. Find the value of:

$$\tan \left\{\frac{1}{2}\right[{\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{2x}{1+x^2}\right)+{\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1-y^2}{1+y^2}\right)\left]\right\},$$

given $\mathrm{\mid }x\mathrm{\mid }<1,y>0,xy<1$

Solution. Use known identities:

$${\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x(\mathrm{\mid }x\mathrm{\mid }<1),$$

and for $y>0$,

$${\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1-y^2}{1+y^2}\right)=2{\tan }^{-1}y\mathrm{.}$$

Hence the bracket is $\frac{1}{2}[2{\tan }^{-1}x+2{\tan }^{-1}y]={\tan }^{-1}x+{\tan }^{-1}y$. Therefore

$$\tan ({\tan }^{-1}x+{\tan }^{-1}y)=\frac{x+y}{1-xy},$$

using the tangent addition formula and the hypothesis $xy<1$. So the value is ${\displaystyle \frac{x+y}{1-xy}}$

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10. Evaluate: ${\sin }^{-1}\text{\hspace{0.17em}⁣}(\sin \frac{2\pi }{3})$.

Solution. $\sin \frac{2\pi }{3}=\frac{\sqrt{3}}{2}$. Principal value of ${\sin }^{-1}$ is in $[-\frac{\pi }{2},\frac{\pi }{2}]$; the angle in that range with sine $\frac{\sqrt{3}}{2}$ is $\frac{\pi }{3}$. Hence

$${\sin }^{-1}\text{\hspace{0.17em}⁣}(\sin \frac{2\pi }{3})=\frac{\pi }{3}\mathrm{}$$

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11. Evaluate: ${\tan }^{-1}\text{\hspace{0.17em}⁣}(\tan \frac{3\pi }{4})$.

Solution. $\tan \frac{3\pi }{4}=-1$. Principal value of ${\tan }^{-1}$ lies in $(-\frac{\pi }{2},\frac{\pi }{2})$; ${\tan }^{-1}(-1)=-\frac{\pi }{4}$. Thus

$${\tan }^{-1}\text{\hspace{0.17em}⁣}(\tan \frac{3\pi }{4})=-\frac{\pi }{4}\mathrm{}$$

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12. Evaluate: $\tan \text{\hspace{0.17em}⁣}({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})$.

Solution. Let $A={\sin }^{-1}\frac{3}{5}$. Then $\sin A=\frac{3}{5}\Rightarrow \cos A=\frac{4}{5}\Rightarrow \tan A=\frac{3}{5}\mathrm{/}\frac{4}{5}=\frac{3}{4}$
Let $B={\cot }^{-1}\frac{3}{2}$. Then $\cot B=\frac{3}{2}\Rightarrow \tan B=\frac{2}{3}$
So

$$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot \frac{2}{3}}=\frac{\frac{9+8}{12}}{1-\frac{6}{12}}=\frac{\frac{17}{12}}{\frac{1}{2}}=\frac{17}{6}\mathrm{}$$

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13. Evaluate (MCQ): ${\cos }^{-1}\text{\hspace{0.17em}⁣}(\cos \frac{7\pi }{6})=$ ?

Options: $\frac{7\pi }{6},\frac{5\pi }{6},\frac{\pi }{3},\frac{\pi }{6}$.

Solution. $\cos \frac{7\pi }{6}=-\frac{\sqrt{3}}{2}$. The principal value range of ${\cos }^{-1}$ is $[0,\pi ]$. The angle in $[0,\pi ]$ with cosine $-\frac{\sqrt{3}}{2}$ is $\frac{5\pi }{6}$. So the answer is $\boxed{{\displaystyle \frac{5\pi }{6}}}$.

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14. Evaluate (MCQ): $\sin \text{\hspace{0.17em}⁣}(\frac{\pi }{3}-{\sin }^{-1}(-\frac{1}{2}))=$ ?

Options: $\frac{1}{2},\frac{1}{3},\frac{1}{4},1$

Solution. ${\sin }^{-1}(-\frac{1}{2})=-\frac{\pi }{6}$. So the argument is $\frac{\pi }{3}-(-\frac{\pi }{6})=\frac{\pi }{2}$. Hence $\sin (\frac{\pi }{2})=1$. So the answer is $\boxed{{\displaystyle 1}}$.

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15. Evaluate (MCQ): ${\tan }^{-1}\sqrt{3}-{\cot }^{-1}(-\sqrt{3})=$ ?

Options: $\pi ,-\frac{\pi }{2},0,2\sqrt{3}\mathrm{}$

Solution. ${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$. For ${\cot }^{-1}(-\sqrt{3})$ take principal range $(0,\pi )$. Solve $\cot \theta =-\sqrt{3}\Rightarrow \theta =5\pi \mathrm{/}6$ince $\cot (5\pi \mathrm{/}6)=-\sqrt{3}$. Thus

$$\frac{\pi }{3}-\frac{5\pi }{6}=-\frac{\pi }{2}\mathrm{}$$

So the value is $\boxed{{\displaystyle -\frac{\pi }{2}}}$.