Q.If $\sin^{-1}\dfrac{x}{5}+\sec^{-1}\dfrac{5}{4}=\dfrac{\pi}{2}$, then $x =$ ____. Choices given: $[1,\ 3,\ 4,\ 5]$

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of \tan^{-1}\left(\frac{1}{x}\right) will be:

   (a) \tan x

   (b) \cot x

   (c) \cot^{-1}\frac{1}{x}

   (d) \cot^{-1} x

   ›Reveal solutionSolution

   $\tan^{-1}(1/x)$ and $\cot^{-1}x$ are the same angle for positive $x$, because tangent and cotangent are reciprocal functions.

   Concept: For $x>0$, the standard identity is

   $\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x$ …

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2. CBSE 2025Set ANNUAL1 markMCQ
   Q.Value of $\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$ is:

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

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

   (c) $\dfrac{1}{\sqrt{3}}$

   (d) $\dfrac{5\pi}{6}$

   ›Reveal solutionSolution

   Use periodicity of $\tan$ to bring the angle into the principal range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ of $\tan^{-1}$.

   Since $\tan$ has period $\pi$: $\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6}-\pi\right) = \tan\dfrac{\pi}{6}$.

   …

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3. CBSE 2025Set A1 mark
   Q.If $-1 \le x \le 1$, then $\sin(\sin^{-1}x) =$ ______.

   ›Reveal solutionSolution

   $\sin$ and $\sin^{-1}$ are inverse functions of each other on $[-1,1]$, so applying one after the other returns the original input.

   For $-1 \le x \le 1$, $\sin^{-1}x$ is defined and gives an angle $\theta \in \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ such that $\sin\theta = x$. Applying $\sin$ to this angle simply recovers $x$: …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The principal value of $\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) + \tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$ is

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

   (b) $\dfrac{\pi}{12}$

   (c) $\dfrac{5\pi}{12}$

   (d) $\dfrac{7\pi}{12}$

   ›Reveal solutionSolution

   Bring each angle into the principal-value range using $\sin^{-1}(\sin\theta)=\pi-\theta$ and $\tan^{-1}(\tan\theta)=\theta-\pi$ for $\theta$ in the second quadrant.

   Term 1: $\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right)$

   The range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. Since $\dfrac{2\pi}{3}\in\left(\dfrac{\pi}{2},\pi\right)$ lies outside this range, use

   $\sin^{-1}(\sin\theta) = \pi-\theta \quad \text{for } \theta\in\left(\tfrac{\pi}{2},\pi\right)$

   $\sin^{-1}\left(\sin\dfrac{2\pi}{3}\right) = \pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}$

   Term 2: $\tan^{-1}\left(\tan\dfrac{3\pi}{4}\right)$

   …

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5. CBSE 2024Set D1 markMCQ
   Q.$x \in [-1, 1],\ \sin^{-1}(-x) =$

   (a) $-\sin^{-1} x$

   (b) $\sin^{-1} x$

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

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

   ›Reveal solutionSolution

   $\sin^{-1}$ is odd: $\sin^{-1}(-x)=-\sin^{-1}x$.

   For $x\in[-1,1]$, the inverse sine is an odd function, so

   …

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6. CBSE 2024Set ANNUAL1 mark
   Q.The value of $\sin^{-1}\left(\sin \dfrac{2\pi}{3}\right)$ is ________.

   ›Reveal solutionSolution

   Since $\dfrac{2\pi}{3}$ is outside the principal range of $\sin^{-1}$, first rewrite $\sin\dfrac{2\pi}{3}$ as the sine of an angle inside $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$.

   $\sin\dfrac{2\pi}{3} = \sin\left(\pi - \dfrac{2\pi}{3}\right) = \sin\dfrac{\pi}{3}$

   …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.The principal value of $\tan^{-1}\left(\tan \dfrac{7\pi}{6}\right)$

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

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

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

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

   ›Reveal solutionSolution

   Reduce the angle into the principal branch $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ using the period $\pi$ of $\tan$.

   $\tan^{-1}(\tan\theta)=\theta$ holds only when $\theta$ already lies in the principal value branch $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.

   Here $\theta=\dfrac{7\pi}{6}$, which does not lie in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Since $\tan$ has period $\pi$:

   $\tan\dfrac{7\pi}{6} = \tan\left(\dfrac{7\pi}{6}-\pi\right) = \tan\dfrac{\pi}{6}$ …

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8. CBSE 2023Set E1 markMCQ
   Q.$\sin\left(\sin^{-1}\dfrac{1}{2}\right) =$

   (a) $1$

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

   (c) $\dfrac{\sqrt{3}}{2}$

   (d) $0$

   ›Reveal solutionSolution

   $\sin\left(\sin^{-1}\dfrac{1}{2}\right)=\dfrac{1}{2}$.

   Since $\dfrac{1}{2}\in[-1,1]$, $\sin(\sin^{-1}x)=x$ holds:

   …

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9. CBSE 2022Set ANNUAL1 markMCQ
   Q.Write the value of $\sin^{-1}\dfrac13 + \cos^{-1}\dfrac13$.

   (a) $0$

   (b) $1$

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

   (d) $\pi$

   ›Reveal solutionSolution

   Use the standard identity $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ for any $x\in[-1,1]$.

   Here $x=\dfrac13\in[-1,1]$, so directly:

   …

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10. CBSE 2022Set ANNUAL1 markMCQ
    Q.If $\sin^{-1}x = \dfrac{\pi}{5}$, then what is the value of $\cos^{-1}x$?

    (a) $\dfrac{9\pi}{10}$

    (b) $\dfrac{7\pi}{10}$

    (c) $\dfrac{5\pi}{10}$

    (d) $\dfrac{3\pi}{10}$

    ›Reveal solutionSolution

    Use $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ to get $\cos^{-1}x$ directly from $\sin^{-1}x$.

    Given $\sin^{-1}x=\dfrac{\pi}{5}$.

    …

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11. CBSE 2022Set ANNUAL1 mark
    Q.If $\sin^{-1}\dfrac{x}{5}+\sec^{-1}\dfrac{5}{4}=\dfrac{\pi}{2}$, then $x =$ ____. Choices given: $[1,\ 3,\ 4,\ 5]$

    ›Reveal solutionSolution

    Rewrite $\sec^{-1}$ in terms of $\cos^{-1}$ and use $\sin^{-1}z+\cos^{-1}z=\dfrac{\pi}{2}$.

    $\sec^{-1}\dfrac54=\cos^{-1}\dfrac45$ (since $\sec^{-1}y=\cos^{-1}\frac1y$).

    Given: $\sin^{-1}\dfrac x5+\cos^{-1}\dfrac45=\dfrac{\pi}{2}$.

    …

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12. CBSE 2022Set ANNUAL1 markMCQ
    Q.$\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right)$ is equal to OR If $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x) = (3-x^3)^{1/3}$, then $f \circ f(x)$ is

    (a) $x^{1/3}$

    (b) $x^3$

    (c) $x$

    (d) None of the above

    (a) $\dfrac{7\pi}{6}$

    (b) $\dfrac{5\pi}{6}$

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

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

    ›Reveal solutionSolution

    $\cos^{-1}(\cos\theta)$ returns the angle in the principal range $[0,\pi]$ having the same cosine; $\tfrac{7\pi}{6}$ lies outside it.

    The principal-value range of $\cos^{-1}$ is $[0,\pi]$, but $\dfrac{7\pi}{6} \notin [0,\pi]$. Compute the cosine:

    $\cos\frac{7\pi}{6} = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos\frac{\pi}{6} = -\frac{\sqrt3}{2}.$

    We need the angle $\theta \in [0,\pi]$ with $\cos\theta = -\dfrac{\sqrt3}{2}$. That angle is

    $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.$

    Hence $\cos^{-1}\left(\cos\dfrac{7\pi}{6}\right) = \dfrac{5\pi}{6}$, option (b).

    …

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