Inverse Trigonometric Functions Class 12 Maths NCERT Solutions

Step-by-step solution

1. 1Let ${\sin }^{-1}\left(-\frac{1}{2}\right)=y$, so $\sin y=-\frac{1}{2}$.
2. 2The principal branch of ${\sin }^{-1}$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.
3. 3Since $\sin \left(-\frac{\pi }{6}\right)=-\frac{1}{2}$, $y=-\frac{\pi }{6}$.

Step-by-step solution

1. 1Let ${\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=y$, so $\cos y=\frac{\sqrt{3}}{2}$.
2. 2The principal branch of ${\cos }^{-1}$ is $[0,\pi ]$.
3. 3Since $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$, $y=\frac{\pi }{6}$.

Step-by-step solution

1. 1Let ${\text{cosec}}^{-1}(2)=y$, so $\sin y=\frac{1}{2}$.
2. 2The principal branch of ${\text{cosec}}^{-1}$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\setminus \{0\}$.
3. 3Since $\sin \frac{\pi }{6}=\frac{1}{2}$, $y=\frac{\pi }{6}$.

Step-by-step solution

1. 1Let ${\tan }^{-1}(-\sqrt{3})=y$, so $\tan y=-\sqrt{3}$.
2. 2The principal branch of ${\tan }^{-1}$ is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
3. 3Since $\tan \left(-\frac{\pi }{3}\right)=-\sqrt{3}$, $y=-\frac{\pi }{3}$.

Step-by-step solution

1. 1Let ${\cos }^{-1}\left(-\frac{1}{2}\right)=y$, so $\cos y=-\frac{1}{2}$.
2. 2In $[0,\pi ]$, the angle with cosine $-1/2$ is $\frac{2\pi }{3}$.

Step-by-step solution

1. 1Let ${\tan }^{-1}(-1)=y$, so $\tan y=-1$.
2. 2In $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, the angle with tangent −1 is $-\frac{\pi }{4}$.

Step-by-step solution

1. 1Let ${\sec }^{-1}\left(\frac{2}{\sqrt{3}}\right)=y$, so $\cos y=\frac{\sqrt{3}}{2}$.
2. 2The principal branch of ${\sec }^{-1}$ is $[0,\pi ]\setminus \left\{\frac{\pi }{2}\right\}$.
3. 3Since $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$, $y=\frac{\pi }{6}$.

Step-by-step solution

1. 1Let ${\cot }^{-1}(\sqrt{3})=y$, so $\cot y=\sqrt{3}$.
2. 2The principal branch of ${\cot }^{-1}$ is $(0,\pi )$.
3. 3Since $\cot \frac{\pi }{6}=\sqrt{3}$, $y=\frac{\pi }{6}$.

Step-by-step solution

1. 1Let ${\cos }^{-1}\left(-\frac{1}{\sqrt{2}}\right)=y$, so $\cos y=-\frac{1}{\sqrt{2}}$.
2. 2In $[0,\pi ]$, the angle with cosine $-\frac{1}{\sqrt{2}}$ is $\frac{3\pi }{4}$.

Step-by-step solution

1. 1Let ${\text{cosec}}^{-1}(-\sqrt{2})=y$, so $\sin y=-\frac{1}{\sqrt{2}}$.
2. 2In $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\setminus \{0\}$, the angle with sine $-\frac{1}{\sqrt{2}}$ is $-\frac{\pi }{4}$.

Step-by-step solution

1. 1${\tan }^{-1}(1)=\frac{\pi }{4}$.
2. 2${\cos }^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi }{3}$ and ${\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi }{6}$.
3. 3Sum: $\frac{\pi }{4}+\frac{2\pi }{3}-\frac{\pi }{6}=\frac{3\pi +8\pi -2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}$.

Step-by-step solution

1. 1${\cos }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$.
2. 2${\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$.
3. 3Sum: $\frac{\pi }{3}+2\cdot \frac{\pi }{6}=\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}$.

Step-by-step solution

1. 1The range of the principal value branch of ${\sin }^{-1}$ is $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.
2. 2So $y={\sin }^{-1}x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

Step-by-step solution

1. 1${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$.
2. 2${\sec }^{-1}(-2)=\frac{2\pi }{3}$ (principal branch $[0,\pi ]\setminus \{\pi /2\}$).
3. 3$\frac{\pi }{3}-\frac{2\pi }{3}=-\frac{\pi }{3}$.

Step-by-step solution

1. 1Let $x=\sin \theta$, so $\theta ={\sin }^{-1}x$ and $\theta \in \left[-\frac{\pi }{6},\frac{\pi }{6}\right]$.
2. 2Then $3x-4x^3=3\sin \theta -4{\sin }^3\theta =\sin 3\theta$.
3. 3Since $3\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, ${\sin }^{-1}(\sin 3\theta )=3\theta =3{\sin }^{-1}x$.

Step-by-step solution

1. 1Let $x=\cos \theta$, so $\theta ={\cos }^{-1}x$ and $\theta \in \left[0,\frac{\pi }{3}\right]$.
2. 2Then $4x^3-3x=4{\cos }^3\theta -3\cos \theta =\cos 3\theta$.
3. 3Since $3\theta \in [0,\pi ]$, ${\cos }^{-1}(\cos 3\theta )=3\theta =3{\cos }^{-1}x$.

Step-by-step solution

1. 1Use ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$ with $a=\frac{2}{11}$, $b=\frac{7}{24}$.
2. 2$\frac{a+b}{1-ab}=\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\cdot \frac{7}{24}}=\frac{\frac{48+77}{264}}{\frac{264-14}{264}}=\frac{125}{250}=\frac{1}{2}$.
3. 3Both angles are in $\left(0,\frac{\pi }{2}\right)$, so the sum formula applies and the result follows.

Step-by-step solution

1. 1First, $2{\tan }^{-1}\frac{1}{2}={\tan }^{-1}\left(\frac{2\cdot \frac{1}{2}}{1-\frac{1}{4}}\right)={\tan }^{-1}\frac{4}{3}$.
2. 2Now ${\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\left(\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\cdot \frac{1}{7}}\right)={\tan }^{-1}\left(\frac{\frac{28+3}{21}}{\frac{21-4}{21}}\right)={\tan }^{-1}\frac{31}{17}$.

Step-by-step solution

1. 1Let $x=\tan \theta$, $\theta \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
2. 2Then $\sqrt{1+x^2}=\sec \theta$ and $\frac{\sec \theta -1}{\tan \theta }=\frac{1-\cos \theta }{\sin \theta }=\tan \frac{\theta }{2}$.
3. 3So the expression is ${\tan }^{-1}\left(\tan \frac{\theta }{2}\right)=\frac{\theta }{2}=\frac{1}{2}{\tan }^{-1}x$

Step-by-step solution

1. 1Let $x=\text{cosec}\theta$ with $\theta \in \left(0,\frac{\pi }{2}\right)$ for $x>1$.
2. 2Then $\sqrt{x^2-1}=\sqrt{{\text{cosec}}^2\theta -1}=\cot \theta$, giving $\frac{1}{\sqrt{x^2-1}}=\tan \theta$.
3. 3For $x>1$: value $=\theta ={\text{cosec}}^{-1}x$.
4. 4For $-1>x$: $\theta \in \left(-\frac{\pi }{2},0\right)$, so the value is $-{\text{cosec}}^{-1}x$.

Step-by-step solution

1. 1$\frac{1-\cos x}{1+\cos x}=\frac{2{\sin }^2\frac{x}{2}}{2{\cos }^2\frac{x}{2}}={\tan }^2\frac{x}{2}$.
2. 2For $x\in (\pi ,2\pi )$, $\frac{x}{2}\in \left(\frac{\pi }{2},\pi \right)$, so $\tan \frac{x}{2}<0$.
3. 3$\sqrt{\frac{1-\cos x}{1+\cos x}}=-\tan \frac{x}{2}=\tan \left(\pi -\frac{x}{2}\right)$.
4. 4Hence the expression $=\pi -\frac{x}{2}$

Step-by-step solution

1. 1Divide numerator and denominator by $\cos x$:
2. 2$\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}=\tan \left(\frac{\pi }{4}-x\right)$.
3. 3For $0<x<\frac{3\pi }{4}$, $\frac{\pi }{4}-x\in \left(-\frac{\pi }{2},\frac{\pi }{4}\right)$, giving $\frac{\pi }{4}-x$.
4. 4For $\frac{3\pi }{4}<x<\pi$, the value is $\frac{5\pi }{4}-x$.

Step-by-step solution

1. 1Let $x=a\sin \theta$, so $\theta ={\sin }^{-1}\frac{x}{a}$ and $|\theta |<\frac{\pi }{2}$.
2. 2$\frac{x}{\sqrt{a^2-x^2}}=\frac{a\sin \theta }{a\cos \theta }=\tan \theta$.
3. 3So the expression $={\tan }^{-1}(\tan \theta )=\theta ={\sin }^{-1}\frac{x}{a}$

Step-by-step solution

1. 1Let $x=a\tan \theta$, so $\theta ={\tan }^{-1}\frac{x}{a}$ and $|\theta |\le \frac{\pi }{6}$.
2. 2$\frac{3a^{2}x-x^3}{a^3-3a{x}^2}=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=\tan 3\theta$.
3. 3Since $3\theta \in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$, the expression $=3\theta =3{\tan }^{-1}\frac{x}{a}$

Step-by-step solution

1. 1${\sin }^{-1}\frac{1}{2}=\frac{\pi }{6}$.
2. 2$2\cos \left(2\cdot \frac{\pi }{6}\right)=2\cos \frac{\pi }{3}=2\cdot \frac{1}{2}=1$.
3. 3${\tan }^{-1}(1)=\frac{\pi }{4}$

Step-by-step solution

1. 1${\tan }^{-1}a+{\cot }^{-1}a=\frac{\pi }{2}$ for all $a\in \mathbb{R}$.
2. 2$\cot \frac{\pi }{2}=0$

Step-by-step solution

1. 1Let $x=\tan \alpha$; then $\frac{2x}{1+x^2}=\sin 2\alpha$, so ${\sin }^{-1}\frac{2x}{1+x^2}=2{\tan }^{-1}x$.
2. 2Let $y=\tan \beta$; then $\frac{1-y^2}{1+y^2}=\cos 2\beta$, so ${\cos }^{-1}\frac{1-y^2}{1+y^2}=2{\tan }^{-1}y$.
3. 3Half the sum: ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$ since $xy<1$.
4. 4Taking tan $=\frac{x+y}{1-xy}$

Step-by-step solution

1. 1Since the sine of the angle is 1, the angle is $\frac{\pi }{2}$.
2. 2${\sin }^{-1}\frac{1}{5}+{\cos }^{-1}x=\frac{\pi }{2}$.
3. 3${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}\frac{1}{5}={\cos }^{-1}\frac{1}{5}$.

Step-by-step solution

1. 1${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$ with $a=\frac{x-1}{x-2}$, $b=\frac{x+1}{x+2}$.
2. 2$a+b=\frac{2x^2-4}{x^2-4}$ and $1-ab=\frac{-3}{x^2-4}$.
3. 3So $\frac{a+b}{1-ab}=-\frac{2x^2-4}{3}=\tan \frac{\pi }{4}=1$ ⟹ $2x^2-4=-3$ ⟹ $x^2=\frac{1}{2}$.

Step-by-step solution

1. 1$\frac{2\pi }{3}\notin \left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.
2. 2$\sin \frac{2\pi }{3}=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$.
3. 3${\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{3}$

Step-by-step solution

1. 1$\frac{3\pi }{4}\notin \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$.
2. 2$\tan \frac{3\pi }{4}=-1$.
3. 3${\tan }^{-1}(-1)=-\frac{\pi }{4}$

Step-by-step solution

1. 1Let ${\sin }^{-1}\frac{3}{5}=\alpha$, so $\tan \alpha =\frac{3}{4}$.
2. 2Let ${\cot }^{-1}\frac{3}{2}=\beta$, so $\tan \beta =\frac{2}{3}$.
3. 3$\tan (\alpha +\beta )=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\cdot \frac{2}{3}}=\frac{\frac{17}{12}}{\frac{1}{2}}=\frac{17}{6}$

Step-by-step solution

1. 1$\cos \frac{7\pi }{6}=-\frac{\sqrt{3}}{2}$.
2. 2The principal angle in $[0,\pi ]$ with this cosine is $\frac{5\pi }{6}$.

Step-by-step solution

1. 1${\sin }^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi }{6}$.
2. 2$\frac{\pi }{3}-\left(-\frac{\pi }{6}\right)=\frac{\pi }{2}$.
3. 3$\sin \frac{\pi }{2}=1$

Step-by-step solution

1. 1${\tan }^{-1}\sqrt{3}=\frac{\pi }{3}$.
2. 2${\cot }^{-1}(-\sqrt{3})=\frac{5\pi }{6}$ (principal branch $(0,\pi )$).
3. 3$\frac{\pi }{3}-\frac{5\pi }{6}=-\frac{\pi }{2}$