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Inverse Trigonometric Functions - ISC Class 12 Mathematics Questions with Answers, Page 2

49 past-paper questions on Inverse Trigonometric Functions from ISC Class 12 Mathematics papers (2027-2017), newest first, in full. Questions 21-40 are on this page, 20 to a page. Tap "Show answer" under a question to see its answer.

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2025 · 6 marks · Short answerOpen: Observe the two graphs, Graph 1 and Graph 2 given below and answer the…
Observe the two graphs, Graph 1 and Graph 2 given below and answer the questions that follow.
Figure for this question
(i)[1.0]
Which one of the graphs represents $y = \sin^{-1} x$?
(ii)[1.0]
Write the domain and range of $y = \sin^{-1} x$.
(iii)[2.0]
Prove that $\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}$.
(iv)[2.0]
Find the value of $\tan^{-1}\left(2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right)$

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2025 · 4 marks · Short answerOpen: Solve: . [Application]
Solve: $\sin^{-1}(x) + \sin^{-1}(1-x) = \cos^{-1} x$. [Application]

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2025 · 2 marks · Short answerOpen: Find the value of , if
Find the value of $\tan^{-1} x - \cot^{-1} x$, if $(\tan^{-1} x)^2 - (\cot^{-1} x)^2 = \frac{5\pi^2}{8}$

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2025 · 6 marks · Short answerOpen: Consider the function Find the range of if . Prove: . Evaluate: . If …
Consider the function $y = \sin^{-1}(2x\sqrt{1 - x^2}), x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]$
(i)[1.0]
Find the range of $y$ if $x \in \left[0, \frac{1}{\sqrt{2}}\right]$.
(ii)[1.0]
Prove: $2\sin^{-1}x = \sin^{-1}(2x\sqrt{1 - x^2})$.

Given: $x = \sin\theta$

To show: $2\sin^{-1}x = \sin^{-1}(2x\sqrt{1 - x^2})$

(iii)[2.0]
Evaluate: $\tan^{-1}\left(2\cos\left(2\sin^{-1}\frac{\sqrt{3}}{2}\right)\right)$.
(iv)[2.0]
If $4\sin^{-1}x + \cos^{-1}x = \pi$, calculate the value of '$x$'.

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2025 · 2 marks · Short answerOpen: If , then find .
If $\tan^{-1}\left(\frac{1}{1+1\cdot 2}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \dots + \tan^{-1}\left(\frac{1}{111}\right) = S$, then find $\tan S$.

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2024 · 4 marks · NumericalOpen: Solve for :
Solve for $x$: $\sin^{-1}\left(\frac{x}{2}\right) + \cos^{-1} x = \frac{\pi}{6}$

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2024 · 1 mark · MCQOpen: The value of is equal to:
The value of $\operatorname{cosec}\left(\sin^{-1}\left(-\frac{1}{2}\right)\right) - \sec\left(\cos^{-1}\left(-\frac{1}{2}\right)\right)$ is equal to:
  • (a)$-4$
  • (b)$0$
  • (c)$-1$
  • (d)$4$

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2024 · 4 marks · DerivationOpen: If , show that
If $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$, show that $x^2 - y^2 - z^2 + 2yz\sqrt{1 - x^2} = 0$

Given: $\sin^{-1} x + \sin^{-1} y + \sin^{-1} z = \pi$

To show: $x^2 - y^2 - z^2 + 2yz\sqrt{1 - x^2} = 0$

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2023 · 1 mark · MCQOpen: The value of is equal to
The value of $\tan^{-1} \sqrt{3} - \sec^{-1}(-2)$ is equal to
  • (a)$\frac{\pi}{3}$
  • (b)$\frac{2\pi}{3}$
  • (c)$-\frac{\pi}{3}$
  • (d)$\frac{\pi}{4}$

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2023 · 4 marks · DerivationOpen: If then prove that
If $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$ then prove that $24x^2 - 23x - 12 = 0$

Given: $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$

To show: $24x^2 - 23x - 12 = 0$

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2023 · 2 marks · Short answerOpen: Solve for :
Solve for $x$: $5\tan^{-1} x + 3\cot^{-1} x = 2\pi$

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2023 · 4 marks · DerivationOpen: If , prove that
If $\cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha$, prove that $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha$

Given: $\cos^{-1} \frac{x}{a} + \cos^{-1} \frac{y}{b} = \alpha$

To show: $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = \sin^2 \alpha$

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2022 · 2 marks · MCQOpen: ,
$\forall x \in \mathbb{R}$, $\cot^{-1}(-x) =$
  • (a)$\pi - \cot^{-1} x$
  • (b)$-\tan^{-1} x$
  • (c)$-\cot^{-1} x$
  • (d)$\pi + \cot^{-1} x$

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2022 · 2 marks · MCQOpen: If , then is:
If $\alpha \le 2\sin^{-1} x + \cos^{-1} x \le \beta$, then $(\alpha, \beta)$ is:
  • (a)$(0, \pi)$
  • (b)$(-\frac{\pi}{2}, \frac{\pi}{2})$
  • (c)$(-\frac{3\pi}{2}, \frac{\pi}{2})$
  • (d)None of the above.

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2022 · 2 marks · MCQOpen: What will be the Principal value of ?
What will be the Principal value of $\operatorname{cosec}^{-1}(-\sqrt{2})$?
  • (a)$\frac{3\pi}{4}$
  • (b)$-\frac{\pi}{6}$
  • (c)$\frac{\pi}{4}$
  • (d)$-\frac{\pi}{4}$

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2021 · 4 marks · DerivationOpen: Prove that .
Prove that $\tan^{-1}\frac{1}{2} = \frac{\pi}{4} - \frac{1}{2} \cos^{-1}\left(\frac{4}{5}\right)$.

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2021 · 1 mark · MCQOpen: Simplified value of is:
Simplified value of $\sin\left[\frac{\pi}{2} - \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$ is:
  • (a)$\frac{1}{2}$
  • (b)$\frac{1}{\sqrt{2}}$
  • (c)$\frac{\sqrt{3}}{2}$
  • (d)$-\frac{\sqrt{3}}{2}$

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2020 · 4 marks · DerivationOpen: If , then prove that .
If $\cos^{-1}\frac{x}{2} + \cos^{-1}\frac{y}{3} = \theta$, then prove that $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$.

Given: $\cos^{-1}\frac{x}{2} + \cos^{-1}\frac{y}{3} = \theta$

To show: $9x^2 - 12xy\cos\theta + 4y^2 = 36\sin^2\theta$

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2020 · 4 marks · NumericalOpen: Evaluate: at .
Evaluate: $\cos\left(2\cos^{-1} x + \sin^{-1} x\right)$ at $x = \frac{1}{5}$.

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2020 · 2 marks · DerivationOpen: Prove that: .
Prove that: $\tan^2(\sec^{-1} 2) + \cot^2(\operatorname{cosec}^{-1} 3) = 11$.

Given: $\tan^2(\sec^{-1} 2) + \cot^2(\operatorname{cosec}^{-1} 3)$

To show: $11$

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