Class 12 Mathematics Chapter 2 Inverse Trigonometric Functions

Key Concept: Principal Value

b) $[0, \pi] – \{\pi/2\}$
[Solution Description]
The principal value branch for the inverse secant function, $\sec^{-1}(x)$, has a domain of $R – (-1, 1)$ and a range of $[0, \pi] – \{\pi/2\}$. This means that the angles it can take are between zero and pi, excluding pi over two.

Key Concept: Inverse Function Domain

b) $[-1, 1]$

[Solution Description] The inverse cosine function, $\cos^{-1}(x)$, is defined for values of $x$ where the cosine function exists, specifically within the interval $[-1, 1]$, which includes all possible cosine values.

Key Concept: Basic Identity

c) $\sin^{-1}(\sqrt{1-x^2}) = 2\sin^{-1}x$
[Solution Description]
The given identity $\sin^{-1}(\sqrt{1-x^2}) = 2\sin^{-1}x$ is applicable when the range for $x$ is $-\frac{1}{2} \leq x \leq \frac{1}{2}$. This matches the condition provided in the question.

Key Concept: Function Notation

c) $\cos^{-1}(x)$

[Solution Description] The inverse of the cosine function is typically denoted as $\cos^{-1}(x)$ or alternatively as $\text{arccos}(x)$. Both notations are correct and commonly used.

Key Concept: Domain Recognition

a) $R – (-1, 1)$
[Solution Description] The domain of the inverse secant function, $\sec^{-1} x$, includes all real numbers except those in the open interval (-1, 1). This is because the secant function itself becomes undefined within this interval.

Key Concept: Domain Identification

d) $[-1, 1]$
[Solution Description] The domain of an inverse cosine function $\cos^{-1} x$ is defined as the set of values for which the function can take inputs, i.e., where the function is defined. For $\cos^{-1} x$, this is $[-1, 1]$.

Key Concept: Principal Value Calculation

b) $-\frac{\pi}{4}$

[Solution Description]

We need to determine the principal value of $\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)$, which lies within the range $[-\pi/2, \pi/2]$ for the inverse sine function.

Knowing that $\sin(\frac{-\pi}{4}) = -\frac{\sqrt{2}}{2}$, the principal value will be $-\frac{\pi}{4}$, as it falls into the required range.

Key Concept: Domain Recognition

c) R
[Solution Description] The function $\tan^{-1} x$ has a domain of all real numbers as it accepts any real number input. Thus, its domain is R.

Key Concept: Basic Identity

c) $\frac{\pi}{2} – x$
[Solution Description] For $ 0 \leq x \leq \frac{\pi}{2} $, $\cos(x)$ can be expressed as $\sqrt{1-\sin^2(x)}$. Therefore, $\sin^{-1}(\cos(x))$ equals $\frac{\pi}{2} – x$ because $\sin^{-1}(\cos(x))$ simplifies to an angle whose sine would result in $cos(x)$.

Key Concept: Range Identification

c) $(-\pi/2, \pi/2)$
[Solution Description] The principal value branch of the inverse tangent function $\tan^{-1} x$ is $(-\pi/2, \pi/2)$. Thus, this interval represents its range.

Key Concept: Function Range

b) $[-\frac{\pi}{2}, \frac{\pi}{2}]$
[Solution Description] The range of the inverse sine function $y = \sin^{-1} x$ is from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ inclusive. Hence, the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Key Concept: Identity Application

a) True
[Solution Description]
We need to prove the identity: $\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}$

Recall that by definition: $\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)$

Substitute this into the identity: $\tan^{-1}(x) + \tan^{-1}\left(\frac{1}{x}\right)$

Using the property for addition of arctangent functions: $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$

When $a = x$ and $b = \frac{1}{x}$ then: $\tan^{-1}\left(\frac{x + \frac{1}{x}}{1 – x \cdot \frac{1}{x}}\right) = \tan^{-1}\left(\frac{x^2 + 1}{0}\right) = \tan^{-1}(\infty) = \frac{\pi}{2}
$

Thus, the identity holds true for all $x > 0$.

Key Concept: Function Notation

c) $\frac{1}{\sin x}$
[Solution Description] The reciprocal of the sine function is represented as $\frac{1}{\sin x}$, which is the cosecant function denoted by $\csc x$.

Key Concept: Basic Definition

c) The function f is both one-one and onto
[Solution Description] For a function $f$ to have an inverse, it must be both one-one and onto.

Key Concept: Basic Domain

a) $[-1, 1]$
[Solution Description] The function $y = \cos^{-1} x$ represents the inverse cosine function, and its domain consists of all values that $x$ can take for which $\cos^{-1}$ is defined. The domain of $\cos^{-1} x$ is $[-1, 1]$ because the cosine function produces values in this range.

Key Concept: Graph Reflection

a) It is a reflection of $y = \cos x$ along the line $y = x$.
[Solution Description]
The graph of $y = \cos^{-1} x$ can be obtained from the original function $y = \cos x$ by reflecting it along the line $y = x$. This transformation interchanges the x and y values, thus converting $\cos x$ to its inverse.

Key Concept: Reflection Property

c) $y = x$
[Solution Description] To find the graph of an inverse trigonometric function like $y = \cos^{-1} x$, we reflect the graph of the original function $y = \cos x$ along the line $y = x$. This property holds for all inverse functions.

Key Concept: Symmetry Identification

a) $y = \sin^{-1} x$
[Solution Description] The graph of an inverse function is a reflection of its original function along the line $y = x$. Hence, the graph of any inverse trigonometric function will be symmetric about this line.

Key Concept: Real-world Application, Multi-step Solution

c) 80 meters
[Solution Description]
Given that the angle of elevation is $\theta = \tan^{-1}(0.8)$, then: $\tan(\theta) = 0.8$

Therefore, expressing the problem in terms of trigonometry: $\tan(\theta) = \frac{\text{height}}{\text{distance}}$

Here, $\text{distance} = 100 \, \text{meters}$.

Solving for height ($h$): $h = 100 \times 0.8 = 80 \, \text{meters}$

Hence, the height of the dish is 80 meters.