Inverse Trigonometric Functions — NCERT Solutions

Given: $\sin^{-1}\left(-\dfrac{1}{2}\right)$

Concept: The principal value branch of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Working:

Let $\sin^{-1}\left(-\dfrac{1}{2}\right) = y$. Then $\sin y = -\dfrac{1}{2}$.

We know that $\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$ and $-\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Answer: The principal value of $\sin^{-1}\left(-\dfrac{1}{2}\right) = \boxed{-\dfrac{\pi}{6}}$.

Given: $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$

Concept: The principal value branch of $\cos^{-1}$ is $[0, \pi]$.

Working:

Let $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = y$. Then $\cos y = \dfrac{\sqrt{3}}{2}$.

We know that $\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$ and $\dfrac{\pi}{6} \in [0, \pi]$.

Answer: The principal value of $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = \boxed{\dfrac{\pi}{6}}$.

Given: $\csc^{-1}(2)$

Concept: The principal value branch of $\csc^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$.

Working:

Let $\csc^{-1}(2) = y$. Then $\csc y = 2$, i.e., $\sin y = \dfrac{1}{2}$.

We know that $\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$, so $\csc\left(\dfrac{\pi}{6}\right) = 2$ and $\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$.

Answer: The principal value of $\csc^{-1}(2) = \boxed{\dfrac{\pi}{6}}$.

Given: $\tan^{-1}(-\sqrt{3})$

Concept: The principal value branch of $\tan^{-1}$ is $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Working:

Let $\tan^{-1}(-\sqrt{3}) = y$. Then $\tan y = -\sqrt{3}$.

We know that $\tan\left(-\dfrac{\pi}{3}\right) = -\sqrt{3}$ and $-\dfrac{\pi}{3} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Answer: The principal value of $\tan^{-1}(-\sqrt{3}) = \boxed{-\dfrac{\pi}{3}}$.

Given: $\cos^{-1}\left(-\dfrac{1}{2}\right)$

Concept: The principal value branch of $\cos^{-1}$ is $[0, \pi]$.

Working:

Let $\cos^{-1}\left(-\dfrac{1}{2}\right) = y$. Then $\cos y = -\dfrac{1}{2}$.

We know that $\cos\left(\dfrac{2\pi}{3}\right) = -\dfrac{1}{2}$ and $\dfrac{2\pi}{3} \in [0, \pi]$.

Answer: The principal value of $\cos^{-1}\left(-\dfrac{1}{2}\right) = \boxed{\dfrac{2\pi}{3}}$.

Given: $\tan^{-1}(-1)$

Concept: The principal value branch of $\tan^{-1}$ is $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Working:

Let $\tan^{-1}(-1) = y$. Then $\tan y = -1$.

We know that $\tan\left(-\dfrac{\pi}{4}\right) = -1$ and $-\dfrac{\pi}{4} \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Answer: The principal value of $\tan^{-1}(-1) = \boxed{-\dfrac{\pi}{4}}$.

Given: $\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right)$

Concept: The principal value branch of $\sec^{-1}$ is $[0, \pi] - \left\{\dfrac{\pi}{2}\right\}$.

Working:

Let $\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = y$. Then $\sec y = \dfrac{2}{\sqrt{3}}$, i.e., $\cos y = \dfrac{\sqrt{3}}{2}$.

We know that $\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$ and $\dfrac{\pi}{6} \in [0, \pi] - \left\{\dfrac{\pi}{2}\right\}$.

Answer: The principal value of $\sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right) = \boxed{\dfrac{\pi}{6}}$.

Given: $\cot^{-1}(\sqrt{3})$

Concept: The principal value branch of $\cot^{-1}$ is $(0, \pi)$.

Working:

Let $\cot^{-1}(\sqrt{3}) = y$. Then $\cot y = \sqrt{3}$.

We know that $\cot\left(\dfrac{\pi}{6}\right) = \sqrt{3}$ and $\dfrac{\pi}{6} \in (0, \pi)$.

Answer: The principal value of $\cot^{-1}(\sqrt{3}) = \boxed{\dfrac{\pi}{6}}$.

Given: $\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)$

Concept: The principal value branch of $\cos^{-1}$ is $[0, \pi]$.

Working:

Let $\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = y$. Then $\cos y = -\dfrac{1}{\sqrt{2}}$.

We know that $\cos\left(\dfrac{3\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}$ and $\dfrac{3\pi}{4} \in [0, \pi]$.

Answer: The principal value of $\cos^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = \boxed{\dfrac{3\pi}{4}}$.

Given: $\csc^{-1}(-\sqrt{2})$

Concept: The principal value branch of $\csc^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$.

Working:

Let $\csc^{-1}(-\sqrt{2}) = y$. Then $\csc y = -\sqrt{2}$, i.e., $\sin y = -\dfrac{1}{\sqrt{2}}$.

We know that $\sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}$ and $-\dfrac{\pi}{4} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \{0\}$.

Answer: The principal value of $\csc^{-1}(-\sqrt{2}) = \boxed{-\dfrac{\pi}{4}}$.

Given: $\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right)$

Working:

Step 1: Find $\tan^{-1}(1)$.
$$\tan\left(\frac{\pi}{4}\right) = 1 \Rightarrow \tan^{-1}(1) = \frac{\pi}{4}$$

Step 2: Find $\cos^{-1}\left(-\dfrac{1}{2}\right)$.
$$\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \Rightarrow \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$$

Step 3: Find $\sin^{-1}\left(-\dfrac{1}{2}\right)$.
$$\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \Rightarrow \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$$

Step 4: Add all three values.
$$\frac{\pi}{4} + \frac{2\pi}{3} + \left(-\frac{\pi}{6}\right) = \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}$$

Answer: $\tan^{-1}(1) + \cos^{-1}\left(-\dfrac{1}{2}\right) + \sin^{-1}\left(-\dfrac{1}{2}\right) = \boxed{\dfrac{3\pi}{4}}$.

Given: $\cos^{-1}\dfrac{1}{2} + 2\sin^{-1}\dfrac{1}{2}$

Working:

Step 1: Find $\cos^{-1}\dfrac{1}{2}$.
$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \Rightarrow \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$$

Step 2: Find $\sin^{-1}\dfrac{1}{2}$.
$$\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \Rightarrow \sin^{-1}\frac{1}{2} = \frac{\pi}{6}$$

Step 3: Compute the expression.
$$\frac{\pi}{3} + 2 \times \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}$$

Answer: $\cos^{-1}\dfrac{1}{2} + 2\sin^{-1}\dfrac{1}{2} = \boxed{\dfrac{2\pi}{3}}$.

Correct Option: (B) $-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$

Justification: The principal value branch of $\sin^{-1}$ is defined as $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, which is a closed interval. Therefore, if $\sin^{-1}x = y$, then $y \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, i.e., $-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$.

Correct Option: (B) $-\dfrac{\pi}{3}$

Working:

Step 1: Find $\tan^{-1}\sqrt{3}$.
$$\tan\left(\frac{\pi}{3}\right) = \sqrt{3} \Rightarrow \tan^{-1}\sqrt{3} = \frac{\pi}{3}$$

Step 2: Find $\sec^{-1}(-2)$.
$$\sec y = -2 \Rightarrow \cos y = -\frac{1}{2} \Rightarrow y = \frac{2\pi}{3} \in [0,\pi]-\left\{\frac{\pi}{2}\right\}$$
$$\therefore \sec^{-1}(-2) = \frac{2\pi}{3}$$

Step 3: Compute.
$$\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}$$

Answer: Option (B) $-\dfrac{\pi}{3}$.

To Prove: $3\sin^{-1}x = \sin^{-1}(3x - 4x^3)$

Proof:

Let $\sin^{-1}x = \theta$, so $x = \sin\theta$.

Since $x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$, we have $\theta \in \left[-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right]$, which means $3\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Now, using the triple angle formula:
$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta = 3x - 4x^3$$

Since $3\theta \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, we can apply $\sin^{-1}$ to both sides:
$$3\theta = \sin^{-1}(3x - 4x^3)$$

Substituting back $\theta = \sin^{-1}x$:
$$3\sin^{-1}x = \sin^{-1}(3x - 4x^3) \qquad \textbf{Hence Proved.}$$

To Prove: $3\cos^{-1}x = \cos^{-1}(4x^3 - 3x)$

Proof:

Let $\cos^{-1}x = \theta$, so $x = \cos\theta$.

Since $x \in \left[\dfrac{1}{2}, 1\right]$, we have $\theta \in \left[0, \dfrac{\pi}{3}\right]$, which means $3\theta \in [0, \pi]$.

Using the triple angle formula:
$$\cos 3\theta = 4\cos^3\theta - 3\cos\theta = 4x^3 - 3x$$

Since $3\theta \in [0, \pi]$, we can apply $\cos^{-1}$ to both sides:
$$3\theta = \cos^{-1}(4x^3 - 3x)$$

Substituting back $\theta = \cos^{-1}x$:
$$3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \qquad \textbf{Hence Proved.}$$

Given: $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$

Working:

Let $x = \tan\theta$, so $\theta = \tan^{-1}x$, where $\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Then:
$$\sqrt{1+x^2} = \sqrt{1+\tan^2\theta} = \sec\theta$$

Substituting:
$$\tan^{-1}\frac{\sec\theta - 1}{\tan\theta} = \tan^{-1}\frac{1 - \cos\theta}{\sin\theta}$$

Using half-angle identities: $1 - \cos\theta = 2\sin^2\dfrac{\theta}{2}$ and $\sin\theta = 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$:
$$= \tan^{-1}\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}} = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2}$$

Substituting back $\theta = \tan^{-1}x$:
$$= \frac{1}{2}\tan^{-1}x$$

Answer: $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x} = \dfrac{1}{2}\tan^{-1}x$.

Given: $\tan^{-1}\left(\sqrt{\dfrac{1-\cos x}{1+\cos x}}\right)$, $0 < x < \pi$

Working:

Using half-angle identities:
$$1 - \cos x = 2\sin^2\frac{x}{2}, \quad 1 + \cos x = 2\cos^2\frac{x}{2}$$

So:
$$\sqrt{\frac{1-\cos x}{1+\cos x}} = \sqrt{\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}}} = \left|\tan\frac{x}{2}\right|$$

Since $0 < x < \pi$, we have $0 < \dfrac{x}{2} < \dfrac{\pi}{2}$, so $\tan\dfrac{x}{2} > 0$.

Therefore:
$$\tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}$$

Answer: $\tan^{-1}\left(\sqrt{\dfrac{1-\cos x}{1+\cos x}}\right) = \dfrac{x}{2}$.

Given: $\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)$

Working:

Divide numerator and denominator by $\cos x$:
$$\tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right)$$

Using the identity $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ with $A = \dfrac{\pi}{4}$ and $B = x$:
$$= \tan^{-1}\left(\tan\left(\frac{\pi}{4} - x\right)\right)$$

Since $-\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$, we have $-\pi < \dfrac{\pi}{4} - x < \dfrac{\pi}{2}$, but more precisely $\dfrac{\pi}{4} - x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ for the given range.

Therefore:
$$= \frac{\pi}{4} - x$$

Answer: $\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right) = \dfrac{\pi}{4} - x$.

Given: $\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}}$, $|x| < a$

Working:

Let $x = a\sin\theta$, so $\theta = \sin^{-1}\dfrac{x}{a}$, where $\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Then:
$$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta$$

Substituting:
$$\tan^{-1}\frac{a\sin\theta}{a\cos\theta} = \tan^{-1}(\tan\theta) = \theta = \sin^{-1}\frac{x}{a}$$

Answer: $\tan^{-1}\dfrac{x}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}$.

Given: $\tan^{-1}\left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right)$

Working:

Let $x = a\tan\theta$, so $\theta = \tan^{-1}\dfrac{x}{a}$.

Since $-\dfrac{a}{\sqrt{3}} < x < \dfrac{a}{\sqrt{3}}$, we have $-\dfrac{1}{\sqrt{3}} < \tan\theta < \dfrac{1}{\sqrt{3}}$, so $\theta \in \left(-\dfrac{\pi}{6}, \dfrac{\pi}{6}\right)$ and $3\theta \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Substituting $x = a\tan\theta$:
$$\frac{3a^2(a\tan\theta) - (a\tan\theta)^3}{a^3 - 3a(a\tan\theta)^2} = \frac{3a^3\tan\theta - a^3\tan^3\theta}{a^3 - 3a^3\tan^2\theta} = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta} = \tan 3\theta$$

Therefore:
$$\tan^{-1}(\tan 3\theta) = 3\theta = 3\tan^{-1}\frac{x}{a}$$

Answer: $\tan^{-1}\left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right) = 3\tan^{-1}\dfrac{x}{a}$.

Given: $\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac{1}{2}\right)\right]$

Working:

Step 1: Find $\sin^{-1}\dfrac{1}{2}$.
$$\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$$

Step 2: Find $2\sin^{-1}\dfrac{1}{2}$.
$$2 \times \frac{\pi}{6} = \frac{\pi}{3}$$

Step 3: Find $\cos\left(\dfrac{\pi}{3}\right)$.
$$\cos\frac{\pi}{3} = \frac{1}{2}$$

Step 4: Compute the full expression.
$$\tan^{-1}\left[2 \times \frac{1}{2}\right] = \tan^{-1}(1) = \frac{\pi}{4}$$

Answer: $\tan^{-1}\left[2\cos\left(2\sin^{-1}\dfrac{1}{2}\right)\right] = \boxed{\dfrac{\pi}{4}}$.