Tag: Inverse Trigonometric Functions Solutions Class 12th Maths NCERT

EXERCISE 2.1 — Solutions

1. Find the principal value of ${\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)$.

Solution. Principal branch of ${\sin }^{-1}$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$.
$\sin \frac{\pi }{6}=\frac{1}{2}$. $\frac{\pi }{6}\in [-\frac{\pi }{2},\frac{\pi }{2}]$.

$$\boxed{{\displaystyle {\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)=\frac{\pi }{6}\mathrm{}}}$$

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2. Find the principal value of ${\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{\sqrt{3}}{2}\right)$.

Solution. Principal branch of ${\cos }^{-1}$ is $[0,\pi ]$.
$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$ and $\frac{\pi }{6}\in [0,\pi ]$.

$$\boxed{{\displaystyle {\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}\mathrm{}}}$$

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3. Find the principal value of ${\mathrm{cosec}}^{-1}(2)$.

Solution. If $y={\mathrm{cosec}}^{-1}2$ then $\mathrm{cosec}y=2\Rightarrow \sin y=\frac{1}{2}$. Principal branch for ${\mathrm{cosec}}^{-1}$ (as used in the book) corresponds to $y\in [-\frac{\pi }{2},\frac{\pi }{2}]$ excluding 0 (or equivalent principal branch placing the value at $\pi \mathrm{/}6$ or $\pi -\pi \mathrm{/}6$; standard choice giving principal value in $[-\frac{\pi }{2},\frac{\pi }{2}]\setminus \{0\}$yields $y=\frac{\pi }{6}$. So

$$\boxed{{\displaystyle {\mathrm{cosec}}^{-1}(2)=\frac{\pi }{6}\mathrm{}}}$$

(Interpretation note: many texts give ${\mathrm{cosec}}^{-1}2=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$ if using a different branch — but the book’s principal branch choice yields $\pi \mathrm{/}6$.)

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4. Find the principal value of ${\tan }^{-1}(\sqrt{3})$.

Solution. $\tan \frac{\pi }{3}=\sqrt{3}$. Principal branch of ${\tan }^{-1}$ is $(-\frac{\pi }{2},\frac{\pi }{2})$, and $\frac{\pi }{3}\in (-\frac{\pi }{2},\frac{\pi }{2})$.

$$\boxed{{\displaystyle {\tan }^{-1}(\sqrt{3})=\frac{\pi }{3}\mathrm{}}}$$

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5. Find the principal value of ${\cos }^{-1}\text{\hspace{0.17em}⁣}(-\frac{1}{2})$.

Solution. $\cos \frac{2\pi }{3}=-\frac{1}{2}$, and $\frac{2\pi }{3}\in [0,\pi ]$. So

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

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6. Find the principal value of ${\tan }^{-1}(-1)$.

Solution. $\tan (-\frac{\pi }{4})=-1$. Principal branch is $(-\frac{\pi }{2},\frac{\pi }{2})$.

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

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7. Find the principal value of ${\sec }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{2}{\sqrt{3}}\right)$.

Solution. $\sec \theta =\frac{2}{\sqrt{3}}$ ⇒ $\cos \theta =\frac{\sqrt{3}}{2}$ ⇒ $\theta =\frac{\pi }{6}$ (principal ${\sec }^{-1}$ branch is $[0,\pi ]\setminus \{\frac{\pi }{2}\}$).

$$\boxed{{\displaystyle {\sec }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi }{6}\mathrm{}}}$$

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8. Find the principal value of ${\cot }^{-1}(\sqrt{3})$.

Solution. $\cot \theta =\sqrt{3}\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}$. Principal branch of ${\cot }^{-1}$ in the book is $(0,\pi )$, and $\frac{\pi }{6}\in (0,\pi )$.

$$\boxed{{\displaystyle {\cot }^{-1}(\sqrt{3})=\frac{\pi }{6}\mathrm{}}}$$

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9. Find the principal value of ${\cos }^{-1}\text{\hspace{0.17em}⁣}(-\frac{1}{2})$.

(This repeats Q5 — same answer.)

Solution. As in Q5,

$$\boxed{{\displaystyle \frac{2\pi }{3}\mathrm{}}}$$

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10. Find the principal value of ${\mathrm{cosec}}^{-1}(-2)$.

Solution. $\mathrm{cosec}y=-2\Rightarrow \sin y=-\frac{1}{2}$. Principal value of ${\sin }^{-1}$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$, and $\sin (-\frac{\pi }{6})=-\frac{1}{2}$. So choose $y=-\frac{\pi }{6}$ in the principal branch for cosec${}^{-1}$.

$$\boxed{{\displaystyle {\mathrm{cosec}}^{-1}(-2)=-\frac{\pi }{6}\mathrm{}}}$$

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11. Evaluate ${\tan }^{-1}(1)+{\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)+{\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)$

Solution.
${\tan }^{-1}(1)=\frac{\pi }{4}$
${\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)=\frac{\pi }{3}$
${\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)=\frac{\pi }{6}$
Sum: $\frac{\pi }{4}+\frac{\pi }{3}+\frac{\pi }{6}=\frac{3\pi +4\pi +2\pi }{12}=\frac{9\pi }{12}=\frac{3\pi }{4}\mathrm{}$

$$\boxed{{\displaystyle \frac{3\pi }{4}\mathrm{}}}$$

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12. Evaluate ${\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)+2{\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)$.

Solution. ${\cos }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)=\frac{\pi }{3}$, ${\sin }^{-1}\text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)=\frac{\pi }{6}$.
So $\frac{\pi }{3}+2\cdot \frac{\pi }{6}=\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}\mathrm{}$

$$\boxed{{\displaystyle \frac{2\pi }{3}\mathrm{}}}$$

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13. If ${\sin }^{-1}x=y$, then which is correct?

Options:
(A) $0\le y\le \pi$
(B) $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$
(C) $0<y<\pi$
(D) $-\frac{\pi }{2}<y<\frac{\pi }{2}$

Solution. By the principal branch definition, ${\sin }^{-1}x$ takes values in $[-\frac{\pi }{2},\frac{\pi }{2}]$.

$$\boxed{{\displaystyle \text{Correct: (B) }-\frac{\pi }{2}\le y\le \frac{\pi }{2}\mathrm{.}}}$$

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14. We evaluate

$${\tan }^{-1}(\sqrt{3})-{\sec }^{-1}(-2)\mathrm{.}$$

Step 1 — principal values:

• ${\tan }^{-1}(\sqrt{3})={\displaystyle \frac{\pi }{3}}$ (principal value of ${\tan }^{-1}$ is $(-\frac{\pi }{2},\frac{\pi }{2})$, and $\tan \frac{\pi }{3}=\sqrt{3}$).

• ${\sec }^{-1}(-2)$: solve $\sec \theta =-2\Rightarrow \cos \theta =-\frac{1}{2}$. The principal value of ${\sec }^{-1}$ is taken in $[0,\pi ]\setminus \{\frac{\pi }{2}\}$. In that interval the solution is $\theta =\frac{2\pi }{3}$ (since $\cos \frac{2\pi }{3}=-\frac{1}{2}$). So ${}^{}$${\sec }^{-1}(-2)=\frac{2\pi }{3}$

Step 2 — subtract:

$$\frac{\pi }{3}-\frac{2\pi }{3}=-\frac{\pi }{3}\mathrm{.}$$

So the value equals $-{\displaystyle \frac{\pi }{3}}$. (Option B.)