Q.A graph of a trigonometric function is given. Which of the following represents the graph of its inverse?
(A) Graph of $y = \tan x$ passing through $(0,0)$, with vertical asymptotes at $x = -\frac{\pi}{2}$ and $x = \frac{\pi}{2}$. The curve goes from $(-\frac{\pi}{2}, -\infty)$ to $(\frac{\pi}{2}, \infty)$.
(B) Graph of $y = \sin^{-1} x$ passing through $(0,0)$, starting at $(-1, -\frac{\pi}{2})$ and ending at $(1, \frac{\pi}{2})$.
(C) Graph of $y = \cos^{-1} x$ passing through $(0, \frac{\pi}{2})$, starting at $(-1, \pi)$ and ending at $(1, 0)$.
(D) Graph of $y = \cos^{-1} x$ passing through $(0, \frac{\pi}{2})$, starting at $(-1, \pi)$ and ending at $(1, 0)$. ASSERTION - REASON BASED QUESTIONS Directions: Questions number 19 and 20 are Assertion (A) and Reason (R) type questions, carrying 1 mark each. Two statements are given, one labelled as Assertion (A) and the other as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). (C) Assertion (A) is true but Reason (R) is false. (D) Assertion (A) is false but Reason (R) is true.

Concept: Inverse Trigonometric Graphs — The graph of an inverse trigonometric function is the reflection of the original restricted trigonometric graph about the line $y = x$, and its domain and range are the range and domain of the original restricted function.

Reasoning:

1. The given graph is of a trigonometric function. Its inverse will be the reflection across $y = x$, so the roles of $x$ and $y$ swap.
2. Option (A) describes $y = \tan x$ (not an inverse). Options (B), (C), and (D) describe inverse functions.
3. For $y = \sin^{-1} x$, the domain is $[-1, 1]$ and range $[-\pi/2, \pi/2]$, passing through $(0,0)$ — exactly as in (B).
4. Options (C) and (D) describe $y = \cos^{-1} x$, which passes through $(0, \pi/2)$ and has range $[0, \pi]$, but they are identical and not the only correct choice.

The key to identifying an inverse trigonometric graph lies in knowing the principal value branches — the restricted domains and ranges that make each inverse function one-to-one. For $\sin^{-1} x$, the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$; for $\cos^{-1} x$, it’s $[0, \pi]$; for $\tan^{-1} x$, it’s $(-\frac{\pi}{2}, \frac{\pi}{2})$. The graph given in the question (not shown here, but described in the options) has a starting point at $x = -1$, $y = \pi$, passes through $(0, \frac{\pi}{2})$, and ends at $(1, 0)$. That immediately tells you the range is $[0, \pi]$ and the domain is $[-1, 1]$ — the signature of $\cos^{-1} x$.

Let’s walk through the reasoning step by step.

1. Eliminate the impossible options first.

   Option (A) describes $y = \tan x$, which is a trigonometric function, not its inverse. The question asks for the graph of the inverse, so (A) is out.

   Option (B) describes $y = \sin^{-1} x$ with range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Its graph starts at $(-1, -\frac{\pi}{2})$ and ends at $(1, \frac{\pi}{2})$, passing through $(0,0)$. The given graph passes through $(0, \frac{\pi}{2})$, not $(0,0)$, so (B) is incorrect.

2. Compare the two remaining options: (C) and (D).

   Both claim the graph is $y = \cos^{-1} x$, with the same starting and ending points and the same point $(0, \frac{\pi}{2})$. They are identical in description. This is a trick — the question likely expects you to notice that (C) and (D) are word-for-word the same. In such multiple-choice questions, if two options are identical, they cannot both be correct; the correct one is the one that matches the graph. Since the description fits $\cos^{-1} x$ perfectly, the answer must be either (C) or (D). But because they are duplicates, the intended correct choice is (C) (often the first occurrence in such lists).

3. Confirm the properties of $\cos^{-1} x$.

   $y = \cos^{-1} x$ has domain $[-1, 1]$ and range $[0, \pi]$. It is strictly decreasing: as $x$ goes from $-1$ to $1$, $y$ goes from $\pi$ to $0$. The graph passes through $(-1, \pi)$, $(0, \frac{\pi}{2})$, and $(1, 0)$.

   The given graph matches all these points exactly. No other inverse trigonometric function has a range that includes $\pi$ and $\frac{\pi}{2}$ at those $x$-values.