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MP Board Class 12 (MPBSE) Inverse Trigonometric Functions — practice questions

Explore 40 free Inverse Trigonometric Functions questions in MP Board Class 12 (MPBSE) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Inverse Trigonometric Functions questions and exercises with answers

Showing questions 1–30 of 40.

The domain of the function $f(x) = \sin^{-1}(x)$ isRead the answer and worked solution →The range of $\cos^{-1}(x)$ isRead the answer and worked solution →For every $x \in [-1, 1]$, the value of $\sin^{-1}(x) + \cos^{-1}(x)$ isRead the answer and worked solution →The principal value of $\tan^{-1}(1)$ isRead the answer and worked solution →$\tan^{-1}\left(\tfrac{1}{2}\right) + \tan^{-1}\left(\tfrac{1}{3}\right)$ equalsRead the answer and worked solution →$\cot^{-1}\left(-\tfrac{1}{\sqrt{3}}\right) = $Read the answer and worked solution →$\sin^{-1}\left(\sin\tfrac{2\pi}{3}\right)$ equalsRead the answer and worked solution →$\cos^{-1}\left(\cos\tfrac{7\pi}{6}\right)$ equalsRead the answer and worked solution →$\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = $Read the answer and worked solution →$\displaystyle \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right)$ equalsRead the answer and worked solution →For $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equalsRead the answer and worked solution →All real $x$ satisfying $\sin^{-1}(x) + \sin^{-1}(1-x) = \cos^{-1}(x)$ areRead the answer and worked solution →The value of $\cos^{-1}(-1)$ is:Read the answer and worked solution →The value of $\sin(\tan^{-1}(3/4))$ is:Read the answer and worked solution →Sort each inverse function into its domainRead the answer and worked solution →Tap every function whose domain is [−1, 1]Read the answer and worked solution →Match each inverse function to its domainRead the answer and worked solution →Match each inverse function to its principal value branchRead the answer and worked solution →Flip and pair: function with its principal rangeRead the answer and worked solution →Which principal ranges sit around zero, and which run 0 to π?Read the answer and worked solution →Tap every function whose principal range has a point removedRead the answer and worked solution →Name the principal value branchRead the answer and worked solution →Domain checkRead the answer and worked solution →Put the reasoning in order: why an inverse needs a restricted domainRead the answer and worked solution →True or false: existence of inversesRead the answer and worked solution →True or false: reading the tableRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Complete the summary tableRead the answer and worked solution →

Worked practice questions

The domain of the function $f(x) = \sin^{-1}(x)$ is

  1. $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
  2. $\mathbb{R}$
  3. $[-1, 1]$
  4. $[0, \pi]$
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Correct answer: C. $[-1, 1]$

$\sin^{-1}$ is the inverse of sine restricted to $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, where sine takes every value in $[-1,1]$ exactly once. So the inverse is defined precisely on $x \in [-1, 1]$. **Option D is the *range* of $\sin^{-1}$, not the domain.**

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The range of $\cos^{-1}(x)$ is

  1. $[0, \pi]$
  2. $\mathbb{R}$
  3. $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$
  4. $(-\pi, \pi)$
Show answer and solution

Correct answer: A. $[0, \pi]$

By the standard principal-value convention, $\cos^{-1}$ returns angles in $[0, \pi]$ — the interval where cosine is monotonically decreasing and hits every value in $[-1,1]$ exactly once. **Option A is the range of $\sin^{-1}$ and $\tan^{-1}$, not $\cos^{-1}$.**

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For every $x \in [-1, 1]$, the value of $\sin^{-1}(x) + \cos^{-1}(x)$ is

  1. $2\pi$
  2. $\tfrac{\pi}{2}$
  3. $\tfrac{\pi}{4}$
  4. $\pi$
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Correct answer: B. $\tfrac{\pi}{2}$

Let $\theta = \sin^{-1}(x)$ so $x = \sin\theta$ with $\theta \in [-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$. Then $\cos\left(\tfrac{\pi}{2} - \theta\right) = \sin\theta = x$, and $\tfrac{\pi}{2}-\theta \in [0,\pi]$ — the range of $\cos^{-1}$. **So $\cos^{-1}(x) = \tfrac{\pi}{2} - \sin^{-1}(x)$, giving the complementary identity $\sin^{-1}(x) + \cos^{-1}(x) = \tfrac{\pi}{2}$.**

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