NCERT Solutions: Class 12 Maths Chapter 2 - Inverse Trigonometric Functions

The principal range of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. Since $\sin\left(-\dfrac{\pi}{6}\right)=-\dfrac{1}{2}$ and $-\dfrac{\pi}{6}$ lies in this range, $\sin^{-1}\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{6}$.

The principal range of $\cos^{-1}x$ is $[0,\pi]$. Since $\cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}$ and $\dfrac{\pi}{6}\in[0,\pi]$, the principal value is $\dfrac{\pi}{6}$.

Let $\cosec^{-1}(2)=\theta$. Then $\cosec\theta=2$, so $\sin\theta=\dfrac{1}{2}$. In the principal range of $\cosec^{-1}x$, the angle with sine $\dfrac{1}{2}$ is $\dfrac{\pi}{6}$. Hence $\cosec^{-1}(2)=\dfrac{\pi}{6}$.

The principal range of $\tan^{-1}x$ is $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Since $\tan\left(-\dfrac{\pi}{3}\right)=-\sqrt{3}$, the principal value is $-\dfrac{\pi}{3}$.

The principal range of $\cos^{-1}x$ is $[0,\pi]$. Since $\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}$ and $\dfrac{2\pi}{3}$ lies in $[0,\pi]$, the required value is $\dfrac{2\pi}{3}$.

Since $\tan\left(-\dfrac{\pi}{4}\right)=-1$ and $-\dfrac{\pi}{4}$ belongs to the principal range of $\tan^{-1}x$, $\tan^{-1}(-1)=-\dfrac{\pi}{4}$.

Let $\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right)=\theta$. Then $\sec\theta=\dfrac{2}{\sqrt{3}}$, so $\cos\theta=\dfrac{\sqrt{3}}{2}$. In the principal range of $\sec^{-1}x$, this gives $\theta=\dfrac{\pi}{6}$.

The principal range of $\cot^{-1}x$ is $(0,\pi)$. Since $\cot\dfrac{\pi}{6}=\sqrt{3}$, the required principal value is $\dfrac{\pi}{6}$.

The principal value of $\cos^{-1}x$ lies in $[0,\pi]$. Since $\cos\dfrac{3\pi}{4}=-\dfrac{1}{\sqrt{2}}$, we get $\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)=\dfrac{3\pi}{4}$.

Let $\cosec^{-1}(-\sqrt{2})=\theta$. Then $\sin\theta=-\dfrac{1}{\sqrt{2}}$. The principal angle satisfying this is $-\dfrac{\pi}{4}$, so $\cosec^{-1}(-\sqrt{2})=-\dfrac{\pi}{4}$.

Using principal values, $\tan^{-1}(1)=\dfrac{\pi}{4}$, $\cos^{-1}\left(-\dfrac{1}{2}\right)=\dfrac{2\pi}{3}$ and $\sin^{-1}\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{6}$. Therefore the sum is $\dfrac{\pi}{4}+\dfrac{2\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{4}+\dfrac{\pi}{2}=\dfrac{3\pi}{4}$.

$\cos^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{3}$ and $\sin^{-1}\left(\dfrac{1}{2}\right)=\dfrac{\pi}{6}$. Hence the value is $\dfrac{\pi}{3}+2\cdot\dfrac{\pi}{6}=\dfrac{2\pi}{3}$.

1. i. $0 \le y \le \pi$
2. ii. $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$
3. iii. $0 \lt y \lt \pi$
4. iv. $-\dfrac{\pi}{2} \lt y \lt \dfrac{\pi}{2}$

By definition, the principal range of $\sin^{-1}x$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. Therefore $y$ must satisfy $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$, option (ii).

1. i. $\pi$
2. ii. $-\dfrac{\pi}{3}$
3. iii. $\dfrac{\pi}{3}$
4. iv. $\dfrac{2\pi}{3}$

$\tan^{-1}\sqrt{3}=\dfrac{\pi}{3}$. Also $\sec^{-1}(-2)=\dfrac{2\pi}{3}$ because $\sec\dfrac{2\pi}{3}=-2$ in the principal range. Therefore $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)=\dfrac{\pi}{3}-\dfrac{2\pi}{3}=-\dfrac{\pi}{3}$, option (ii).

Let $\theta=\sin^{-1}x$. Since $x\in\left[-\dfrac{1}{2},\dfrac{1}{2}\right]$, we have $\theta\in\left[-\dfrac{\pi}{6},\dfrac{\pi}{6}\right]$, so $3\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, the principal range of $\sin^{-1}$. Now $\sin3\theta=3\sin\theta-4\sin^3\theta=3x-4x^3$. Hence $3\theta=\sin^{-1}(3x-4x^3)$, i.e. $3\sin^{-1}x=\sin^{-1}(3x-4x^3)$.

Let $\theta=\cos^{-1}x$. Since $x\in\left[\dfrac{1}{2},1\right]$, $\theta\in\left[0,\dfrac{\pi}{3}\right]$, so $3\theta\in[0,\pi]$, the principal range of $\cos^{-1}$. Using $\cos3\theta=4\cos^3\theta-3\cos\theta$, we get $\cos3\theta=4x^3-3x$. Therefore $3\theta=\cos^{-1}(4x^3-3x)$, proving the result.

Let $\theta=\tan^{-1}x$. Then $\tan\theta=x$ and $\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Using $\tan\dfrac{\theta}{2}=\dfrac{\sqrt{1+x^2}-1}{x}$, the given expression becomes $\tan^{-1}\left(\tan\dfrac{\theta}{2}\right)$. Since $\dfrac{\theta}{2}\in\left(-\dfrac{\pi}{4},\dfrac{\pi}{4}\right)$, this equals $\dfrac{\theta}{2}=\dfrac{1}{2}\tan^{-1}x$.

For $0\lt x\lt\pi$, $0\lt\dfrac{x}{2}\lt\dfrac{\pi}{2}$ and $\tan\dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$. Therefore the expression is $\tan^{-1}\left(\tan\dfrac{x}{2}\right)=\dfrac{x}{2}$.

We have $\dfrac{\cos x-\sin x}{\cos x+\sin x}=\tan\left(\dfrac{\pi}{4}-x\right)$. Since $-\dfrac{\pi}{4}\lt x\lt\dfrac{3\pi}{4}$, it follows that $-\dfrac{\pi}{2}\lt\dfrac{\pi}{4}-x\lt\dfrac{\pi}{2}$. This lies in the principal range of $\tan^{-1}$. Hence the expression equals $\dfrac{\pi}{4}-x$.

Let $\theta=\sin^{-1}\left(\dfrac{x}{a}\right)$. Since $|x|\lt a$, $\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Also $\sin\theta=\dfrac{x}{a}$ and $\cos\theta=\dfrac{\sqrt{a^2-x^2}}{a}$, so $\tan\theta=\dfrac{x}{\sqrt{a^2-x^2}}$. Therefore the expression is $\tan^{-1}(\tan\theta)=\theta=\sin^{-1}\left(\dfrac{x}{a}\right)$.

Let $\theta=\tan^{-1}\left(\dfrac{x}{a}\right)$. Then $\tan\theta=\dfrac{x}{a}$. The condition $-\dfrac{a}{\sqrt{3}}\lt x\lt\dfrac{a}{\sqrt{3}}$ gives $-\dfrac{\pi}{6}\lt\theta\lt\dfrac{\pi}{6}$, so $3\theta$ lies in the principal range of $\tan^{-1}$. Now $\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=\dfrac{3a^2x-x^3}{a^3-3ax^2}$. Hence the expression equals $3\theta=3\tan^{-1}\left(\dfrac{x}{a}\right)$.

$\sin^{-1}\dfrac{1}{2}=\dfrac{\pi}{6}$. Hence $2\sin^{-1}\dfrac{1}{2}=\dfrac{\pi}{3}$ and $2\cos\dfrac{\pi}{3}=2\cdot\dfrac{1}{2}=1$. Therefore the expression is $\tan^{-1}(1)=\dfrac{\pi}{4}$.

Since $|x|\lt1$, $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)=2\tan^{-1}x$. Since $y\gt0$, $\cos^{-1}\left(\dfrac{1-y^2}{1+y^2}\right)=2\tan^{-1}y$. Therefore the given expression is $\tan(\tan^{-1}x+\tan^{-1}y)$. Using $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, and $xy\lt1$, its value is $\dfrac{x+y}{1-xy}$.

$\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$. The principal value of $\sin^{-1}\left(\dfrac{\sqrt3}{2}\right)$ is $\dfrac{\pi}{3}$ because $\dfrac{\pi}{3}\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. Hence the value is $\dfrac{\pi}{3}$.

$\tan\dfrac{3\pi}{4}=-1$. The principal value of $\tan^{-1}(-1)$ is $-\dfrac{\pi}{4}$, so $\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)=-\dfrac{\pi}{4}$.

Let $A=\sin^{-1}\dfrac{3}{5}$. Then $\tan A=\dfrac{3}{4}$. Let $B=\cot^{-1}\dfrac{3}{2}$. Then $\tan B=\dfrac{2}{3}$. Thus $\tan(A+B)=\dfrac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot\frac{2}{3}}=\dfrac{\frac{17}{12}}{\frac{1}{2}}=\dfrac{17}{6}$.

1. i. $\dfrac{7\pi}{6}$
2. ii. $\dfrac{5\pi}{6}$
3. iii. $\dfrac{\pi}{3}$
4. iv. $\dfrac{\pi}{6}$

$\cos\dfrac{7\pi}{6}=-\dfrac{\sqrt3}{2}$. In the principal range $[0,\pi]$ of $\cos^{-1}x$, the angle with cosine $-\dfrac{\sqrt3}{2}$ is $\dfrac{5\pi}{6}$. Hence the answer is option (ii).

1. i. $\dfrac{1}{2}$
2. ii. $\dfrac{1}{3}$
3. iii. $\dfrac{1}{4}$
4. iv. $1$

$\sin^{-1}\left(-\dfrac{1}{2}\right)=-\dfrac{\pi}{6}$. Therefore $\sin\left(\dfrac{\pi}{3}-\sin^{-1}\left(-\dfrac{1}{2}\right)\right)=\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{2}=1$. Hence option (iv) is correct.

1. i. $\pi$
2. ii. $-\dfrac{\pi}{2}$
3. iii. $0$
4. iv. $2\sqrt{3}$

$\tan^{-1}\sqrt3=\dfrac{\pi}{3}$. Since $\cot\dfrac{5\pi}{6}=-\sqrt3$ and $\cot^{-1}x$ has principal range $(0,\pi)$, $\cot^{-1}(-\sqrt3)=\dfrac{5\pi}{6}$. Hence the expression is $\dfrac{\pi}{3}-\dfrac{5\pi}{6}=-\dfrac{\pi}{2}$, option (ii).

$\cos\dfrac{13\pi}{6}=\cos\left(2\pi+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$. Since $\dfrac{\pi}{6}$ lies in the principal range $[0,\pi]$ of $\cos^{-1}x$, the value is $\dfrac{\pi}{6}$.

$\tan\dfrac{7\pi}{6}=\tan\left(\pi+\dfrac{\pi}{6}\right)=\tan\dfrac{\pi}{6}=\dfrac{1}{\sqrt3}$. The principal value of $\tan^{-1}\left(\dfrac{1}{\sqrt3}\right)$ is $\dfrac{\pi}{6}$.

Let $\theta=\sin^{-1}\dfrac{3}{5}$. Then $\sin\theta=\dfrac{3}{5}$ and $\cos\theta=\dfrac{4}{5}$, so $\tan\theta=\dfrac{3}{4}$. Thus $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2\cdot\frac34}{1-\frac{9}{16}}=\dfrac{24}{7}$. Since $2\theta$ is in the principal range of $\tan^{-1}$, $2\theta=\tan^{-1}\dfrac{24}{7}$. Hence the result follows.

Let $A=\sin^{-1}\dfrac{8}{17}$ and $B=\sin^{-1}\dfrac{3}{5}$. Then $\tan A=\dfrac{8}{15}$ and $\tan B=\dfrac{3}{4}$. Therefore $\tan(A+B)=\dfrac{\frac{8}{15}+\frac34}{1-\frac{8}{15}\cdot\frac34}=\dfrac{\frac{77}{60}}{\frac35}=\dfrac{77}{36}$. Since $A+B$ is acute and lies in the principal range of $\tan^{-1}$, $A+B=\tan^{-1}\dfrac{77}{36}$.

Let $A=\cos^{-1}\dfrac{4}{5}$ and $B=\cos^{-1}\dfrac{12}{13}$. Then $\cos A=\dfrac45$, $\sin A=\dfrac35$, $\cos B=\dfrac{12}{13}$ and $\sin B=\dfrac5{13}$. Hence $\cos(A+B)=\cos A\cos B-\sin A\sin B=\dfrac{48}{65}-\dfrac{15}{65}=\dfrac{33}{65}$. Since $A+B\in[0,\pi]$, $A+B=\cos^{-1}\dfrac{33}{65}$.

Let $A=\cos^{-1}\dfrac{12}{13}$ and $B=\sin^{-1}\dfrac{3}{5}$. Then $\sin A=\dfrac5{13}$, $\cos A=\dfrac{12}{13}$, $\sin B=\dfrac35$ and $\cos B=\dfrac45$. Hence $\sin(A+B)=\sin A\cos B+\cos A\sin B=\dfrac{5}{13}\cdot\dfrac45+\dfrac{12}{13}\cdot\dfrac35=\dfrac{20+36}{65}=\dfrac{56}{65}$. The sum is acute, so it lies in the principal range of $\sin^{-1}$, and $A+B=\sin^{-1}\dfrac{56}{65}$.

Let $A=\sin^{-1}\dfrac{5}{13}$ and $B=\cos^{-1}\dfrac{3}{5}$. Then $\tan A=\dfrac{5}{12}$ and $\tan B=\dfrac{4}{3}$. Therefore $\tan(A+B)=\dfrac{\frac{5}{12}+\frac43}{1-\frac{5}{12}\cdot\frac43}=\dfrac{\frac{7}{4}}{\frac49}=\dfrac{63}{16}$. Since $A+B$ lies in the principal range of $\tan^{-1}$, $A+B=\tan^{-1}\dfrac{63}{16}$.

Let $\theta=\tan^{-1}\sqrt{x}$. Since $x\in[0,1]$, $\theta\in\left[0,\dfrac{\pi}{4}\right]$, so $2\theta\in\left[0,\dfrac{\pi}{2}\right]$. Now $\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-x}{1+x}$. Hence $2\theta=\cos^{-1}\left(\dfrac{1-x}{1+x}\right)$, and therefore $\theta=\dfrac12\cos^{-1}\left(\dfrac{1-x}{1+x}\right)$.

Let the fraction inside $\cot^{-1}$ be $T$. Rationalising, $T=\dfrac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^2}{(1+\sin x)-(1-\sin x)}=\dfrac{2+2\sqrt{1-\sin^2x}}{2\sin x}$. Since $x\in\left(0,\dfrac{\pi}{4}\right)$, $\cos x\gt0$, so $T=\dfrac{1+\cos x}{\sin x}=\cot\dfrac{x}{2}$. As $\dfrac{x}{2}\in\left(0,\dfrac{\pi}{8}\right)$, this lies in the principal range of $\cot^{-1}$. Hence the value is $\dfrac{x}{2}$.

Put $x=\cos2\theta$. Since $\dfrac{1}{\sqrt2}\le x\le1$, we have $0\le\theta\le\dfrac{\pi}{8}$. Then $\sqrt{1+x}=\sqrt{1+\cos2\theta}=\sqrt2\cos\theta$ and $\sqrt{1-x}=\sqrt{1-\cos2\theta}=\sqrt2\sin\theta$. Thus the fraction becomes $\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\tan\left(\dfrac{\pi}{4}-\theta\right)$. Therefore the left side is $\dfrac{\pi}{4}-\theta=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x$.

Taking tangent on both sides, $\tan(2\tan^{-1}(\cos x))=\tan(\tan^{-1}(2\cosec x))$. Hence $\dfrac{2\cos x}{1-\cos^2x}=2\cosec x$. Since $1-\cos^2x=\sin^2x$, this gives $\dfrac{2\cos x}{\sin^2x}=\dfrac{2}{\sin x}$. As $\sin x\ne0$, we get $\cos x=\sin x$, so $\tan x=1$. Therefore $x=n\pi+\dfrac{\pi}{4}$, $n\in\mathbb{Z}$.

Let $\theta=\tan^{-1}x$. Since $x\gt0$, $\theta\in\left(0,\dfrac{\pi}{2}\right)$. The right side is positive, so the left side must be positive; hence $\dfrac{1-x}{1+x}\gt0$ and $0\lt x\lt1$. Taking tangent on both sides gives $\dfrac{1-x}{1+x}=\tan\dfrac{\theta}{2}$. Using $\tan\theta=\dfrac{2t}{1-t^2}$ with $t=\tan\dfrac{\theta}{2}$, we get $x=\dfrac{2\frac{1-x}{1+x}}{1-\left(\frac{1-x}{1+x}\right)^2}=\dfrac{1-x^2}{2x}$. Hence $2x^2=1-x^2$, so $3x^2=1$. Since $x\gt0$, $x=\dfrac1{\sqrt3}$.

1. i. $\dfrac{x}{\sqrt{1-x^2}}$
2. ii. $\dfrac{1}{\sqrt{1-x^2}}$
3. iii. $\dfrac{1}{\sqrt{1+x^2}}$
4. iv. $\dfrac{x}{\sqrt{1+x^2}}$

Let $\theta=\tan^{-1}x$. Then $\tan\theta=x=\dfrac{x}{1}$. In a right triangle, the opposite side can be taken as $x$, the adjacent side as $1$, and the hypotenuse as $\sqrt{1+x^2}$. Hence $\sin\theta=\dfrac{x}{\sqrt{1+x^2}}$. Therefore $\sin(\tan^{-1}x)=\dfrac{x}{\sqrt{1+x^2}}$, option (iv).

1. i. $0,\ \dfrac{1}{2}$
2. ii. $1,\ \dfrac{1}{2}$
3. iii. $0$
4. iv. $\dfrac{1}{2}$

For both inverse sine terms to be defined with $x$ real, $0\le x\le1$. Then $\sin^{-1}x\ge0$, so $\sin^{-1}(1-x)=\dfrac{\pi}{2}+2\sin^{-1}x\ge\dfrac{\pi}{2}$. But $\sin^{-1}(1-x)\le\dfrac{\pi}{2}$. Therefore equality is possible only when $\sin^{-1}x=0$, i.e. $x=0$. Checking, $\sin^{-1}(1)-0=\dfrac{\pi}{2}$. Hence option (iii) is correct.