Q.(a) Evaluate $\sin^{-1}\left(\sin\frac{3\pi}{4}\right) + \cos^{-1}(\cos\pi) + \tan^{-1}(1)$.

Showing the 12 most recent of 54 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right) =$

   (a) $\pi$

   (b) $\frac{2\pi}{3}$

   (c) $\frac{3\pi}{4}$

   (d) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   The sum equals $\frac{3\pi}{4}$.

   Evaluate each principal value:

   • $\tan^{-1}(1) = \frac{\pi}{4}$.
   • $\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}$ (range $[0,\pi]$).
   • $\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}$ (range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$).

   Add, using LCD 12: …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the value of $\cos^{-1}\left(\cos \frac{13\pi}{6}\right)$.

   (a) $\frac{13\pi}{6}$

   (b) $\frac{7\pi}{6}$

   (c) $\frac{5\pi}{6}$

   (d) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)=\frac{\pi}{6}$.

   The principal value branch of $\cos^{-1}$ is $[0,\pi]$. To evaluate $\cos^{-1}\left(\cos\frac{13\pi}{6}\right)$ we must first reduce $\frac{13\pi}{6}$ to an angle whose cosine is the same and which lies in $[0,\pi]$.

   $\frac{13\pi}{6}=2\pi+\frac{\pi}{6}$

   Since cosine has period $2\pi$:

   $\cos\frac{13\pi}{6}=\cos\left(2\pi+\frac{\pi}{6}\right)=\cos\frac{\pi}{6}$

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.The principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is

   (a) $\pi/4$

   (b) $\pi/6$

   (c) $\pi/3$

   (d) $\pi/2$

   ›Reveal solutionSolution

   The principal value branch of $\sin^{-1}$ is $[-\pi/2,\pi/2]$; the angle in this range whose sine is $1/\sqrt2$ is $\pi/4$.

   We need $\theta \in [-\pi/2,\pi/2]$ such that $\sin\theta = \dfrac{1}{\sqrt2}$.

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.sin⁻¹(sin(2π/3)) is equal to

   (a) 2π/3

   (b) π/3

   (c) −π/3

   (d) None of the above

   ›Reveal solutionSolution

   $2\pi/3$ lies outside the principal range $[-\pi/2,\pi/2]$ of $\sin^{-1}$, so we must find the angle inside that range with the same sine value.

   $\sin(2\pi/3) = \sin(\pi-\pi/3) = \sin(\pi/3) = \dfrac{\sqrt3}{2}$.

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is

   (a) $-\frac{\pi}{4}$

   (b) $\frac{\pi}{3}$

   (c) $\frac{\pi}{6}$

   (d) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   $\sin^{-1}\dfrac{1}{\sqrt2}=\dfrac\pi4$.

   The principal value lies in $\left[-\dfrac\pi2,\dfrac\pi2\right]$. Since $\sin\dfrac\pi4=\dfrac{1}{\sqrt2}$, th …

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6. CBSE 2026Set ANNUAL1 mark
   Q.Write the answer in one word/sentence: Write the principal value branches (Range) of $\sin^{-1} x$.

   ›Reveal solutionSolution

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

   To make sine invertible it is restricted to $\left[-\dfrac\pi2,\dfrac\pi2\right]$, on which it is one-one and onto $[-1,1]$; hence this is the pr …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\sin\left[\dfrac{\pi}{2}-\sin^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)\right]$ is equal to

   (a) $1$

   (b) $\dfrac{1}{3}$

   (c) $-1$

   (d) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   $\sin\!\left[\tfrac{\pi}{2}-\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right]=\cos\!\left(\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right)=\tfrac12$.

   Step 1: $\sin\left(\tfrac{\pi}{2}-\theta\right)=\cos\theta$, so the expression equals $\cos\!\left(\sin^{-1}\!\left(-\tfrac{\sqrt3}{2}\right)\right)$.

   …

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8. CBSE 20251 markMCQ
   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.

   ›Reveal solutionSolution

   The given graph shows a curve that starts at $(-1, \pi)$, passes through $(0, \frac{\pi}{2})$, and ends at $(1, 0)$ — this is exactly the principal branch of $y = \cos^{-1} x$. The correct option is (C).

   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 $\cos^{-1} x$. …

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9. CBSE 2025Set 65/2/11 markMCQ
   Q.If $y = \sin^{-1}x$, $-1 \le x \le 0$, then the range of $y$ is: (A) $\left(-\dfrac{\pi}{2}, 0\right)$ (B) $\left[-\dfrac{\pi}{2}, 0\right]$ (C) $\left[-\dfrac{\pi}{2}, 0\right)$ (D) $\left(-\dfrac{\pi}{2}, 0\right]$

   ›Reveal solutionSolution

   For the inverse sine function, the principal value range is $[-\pi/2, \pi/2]$. When $x$ is restricted to $[-1, 0]$, $y$ takes values from $-\pi/2$ up to $0$, including both endpoints. The correct answer is (B).

   The key to this problem is understanding what "principal value" means for inverse trigonometric functions. Unlike regular sine, which is periodic and not one-to-one, $\sin^{-1}x$ (also written as $\arcsin x$) is defined as the inverse of the sine function only on a carefully chosen interval where sine is one-to-one. That interval is $[-\pi/2, \pi/2]$.

   So by definition, for any $x$ in $[-1, 1]$, the value $y = \sin^{-1}x$ is always the unique angle in $[-\pi/2, \pi/2]$ whose sine is $x$. This is the principal value branch — it's not a choice; it's the definition.

   Now the question gives you a further restriction: $x$ is only between $-1$ and $0$. You're being asked: as $x$ runs through that half of the domain, what part of the principal range does $y$ cover?

   Let's work through it.

   1. Recall the principal range of $\sin^{-1}x$.

      The output $y$ always lies in $[-\pi/2, \pi/2]$. That's the full range for the full domain $[-1, 1]$.

   2. Identify the endpoints for $x$ in $[-1, 0]$.

      • When $x = -1$, $y = \sin^{-1}(-1)$. What angle in $[-\pi/2, \pi/2]$ has sine equal to $-1$? That's $-\pi/2$.
      • When $x = 0$, $y = \sin^{-1}(0)$. The angle in $[-\pi/2, \pi/2]$ with sine $0$ is $0$.

   3. Check monotonicity.

      The function $\sin^{-1}x$ is strictly increasing on $[-1, 1]$. So as $x$ increases from $-1$ to $0$, $y$ increases from $-\pi/2$ to $0$. Since the function is continuous and strictly increasing, it hits every value between $-\pi/2$ and $0$.

   4. Are the endpoints included? …

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10. CBSE 2025Set 65/4/11 markMCQ
    Q.The principal value of $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$ is : (A) $-\frac{2\pi}{3}$ (B) $-\frac{\pi}{3}$ (C) $\frac{\pi}{3}$ (D) $\frac{2\pi}{3}$

    ›Reveal solutionSolution

    To find the principal value of $\sin^{-1}(\sin \theta)$, we must ensure the angle $\theta$ lies within the principal value range of $\sin^{-1}(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. By adjusting the given angle $-\frac{10\pi}{3}$ to an equivalent angle within this range, we find the principal value is $\boxed{\frac{\pi}{3}}$.

    The problem asks for the principal value of $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right)$. This involves understanding the definition of the inverse sine function and its principal value branch.

    The inverse sine function, $\sin^{-1}(x)$ (also written as $\arcsin(x)$), gives an angle whose sine is $x$. For $\sin^{-1}(x)$ to be a function, its range must be restricted. By convention, the principal value branch of $\sin^{-1}(x)$ is defined such that its output angle lies in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

    This means that for an expression like $\sin^{-1}(\sin \theta)$, the result is not always simply $\theta$. It is $\theta$ only if $\theta$ itself is already within the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. If $\theta$ is outside this range, we need to find an equivalent angle $\alpha$ such that $\sin \alpha = \sin \theta$ and $\alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, $\sin^{-1}(\sin \theta) = \sin^{-1}(\sin \alpha) = \alpha$.

    Let's apply this concept step-by-step:

    1. Identify the principal value range for $\sin^{-1}(x)$:

       The principal value of $\sin^{-1}(x)$ must lie in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. This is equivalent to angles from $-90^\circ$ to $90^\circ$.

    2. Analyze the inner angle:

       The given angle inside the sine function is $-\frac{10\pi}{3}$.

       We need to evaluate $\sin\left(-\frac{10\pi}{3}\right)$.

       To simplify this, we can add or subtract multiples of $2\pi$ (a full rotation) to find a coterminal angle that is easier to work with.

       $-\frac{10\pi}{3} = -\frac{10\pi}{3} + 4\pi$ (since $4\pi = \frac{12\pi}{3}$)

       $= \frac{-10\pi + 12\pi}{3} = \frac{2\pi}{3}$.

       So, $\sin\left(-\frac{10\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right)$.

       Watch out

       A common mistake is to directly write $\sin^{-1}\left(\sin\left(-\frac{10\pi}{3}\right)\right) = -\frac{10\pi}{3}$. This is incorrect because $-\frac{10\pi}{3}$ (which is $-600^\circ$) is not in the principal value range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (which is $[-90^\circ, 90^\circ]$).

    3. Find an equivalent angle within the principal value range:

       Now we need to find the principal value of $\sin^{-1}\left(\sin\left(\frac{2\pi}{3}\right)\right)$. …

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11. CBSE 2025Set X11 markMCQ
    Q.The principal value of $\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)$ is

    (a) $\dfrac{\pi}{2}$

    (b) $\dfrac{\pi}{3}$

    (c) $\dfrac{\pi}{4}$

    (d) $\dfrac{\pi}{6}$

    ›Reveal solutionSolution

    Principal value of an inverse-sine — correct option is (c). …

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12. CBSE 2025Set E1 markMCQ
    Q.$\sin\left(\sin^{-1}\frac{2\pi}{3}\right) + \tan^{-1}\left(\tan\frac{3\pi}{4}\right) =$

    (a) $\frac{17\pi}{12}$

    (b) $\frac{5}{12}\pi$

    (c) $\frac{\pi}{12}$

    (d) $-\frac{\pi}{12}$

    ›Reveal solutionSolution

    Apply the inverse cancellation and reduce the second term to its principal range; result $\frac{5\pi}{12}$.

    First term: Using $\sin(\sin^{-1}\theta)=\theta$ as intended by the paper, $\sin\left(\sin^{-1}\frac{2\pi}{3}\right)=\frac{2\pi}{3}$. (Strictly, $\frac{2\pi}{3}>1$ is outside the domain of $\sin^{-1}$, but the question intends the direct cancellation.)

    …

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