Q.$\cos^{-1}\left(-\frac{1}{2}\right) =$

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) =$

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

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

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

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

   ›Reveal solutionSolution

   $\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$.

   The range of $\cos^{-1}$ is $[0,\pi]$, but $\frac{7\pi}{6}\notin[0,\pi]$, so we cannot just cancel.

   First find the actual cosine value:

   $\cos\frac{7\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}.$

   …

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2. CBSE 2025Set E1 markMCQ
   Q.$\cos^{-1}\left(\cos\frac{8\pi}{5}\right) =$

   (a) $\frac{8\pi}{5}$

   (b) $\frac{2\pi}{5}$

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

   (d) $\frac{3\pi}{5}$

   ›Reveal solutionSolution

   Reduce the angle to the principal range $[0,\pi]$ of $\cos^{-1}$; the answer is $\frac{2\pi}{5}$.

   The angle $\frac{8\pi}{5}$ is not in $[0,\pi]$, so we cannot just cancel. Use $\cos(2\pi-\theta)=\cos\theta$:

   $\cos\frac{8\pi}{5}=\cos\left(2\pi-\frac{8\pi}{5}\right)=\cos\frac{2\pi}{5}.$

   …

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3. CBSE 2025Set E1 markMCQ
   Q.$\sin\left\{\sin^{-1}\frac{1}{5} + \cos^{-1}x\right\} = 1,\ \Rightarrow x =$

   (a) $1$

   (b) $0$

   (c) $\frac{4}{5}$

   (d) $\frac{1}{5}$

   ›Reveal solutionSolution

   Set the bracket equal to $\frac{\pi}{2}$ and solve; $x=\frac15$.

   Since $\sin\theta=1$ only when $\theta=\frac{\pi}{2}$,

   $\sin^{-1}\frac15+\cos^{-1}x=\frac{\pi}{2}.$

   …

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4. CBSE 2025Set E1 markMCQ
   Q.$\cos^{-1}x + \sec^{-1}\frac{1}{x} =$

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

   (b) $\cos^{-1}(2x^2 - 1)$

   (c) $\cos^{-1}(1 - 2x^2)$

   (d) $\cos^{-1}2x$

   ›Reveal solutionSolution

   $\sec^{-1}\frac1x = \cos^{-1}x$, so the sum is $2\cos^{-1}x$, which equals $\cos^{-1}(2x^2-1)$.

   Use $\sec^{-1}(y) = \cos^{-1}\frac{1}{y}$, so

   $\sec^{-1}\frac{1}{x} = \cos^{-1}x.$

   Therefore

   $\cos^{-1}x + \sec^{-1}\frac{1}{x} = 2\cos^{-1}x.$ …

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5. CBSE 2024Set D1 markMCQ
   Q.$\cos^{-1}\left(-\frac{1}{2}\right) =$

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

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

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

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

   ›Reveal solutionSolution

   The principal value lies in $[0,\pi]$: $\cos^{-1}(-\tfrac12)=\tfrac{2\pi}{3}$.

   The range of $\cos^{-1}$ is $[0,\pi]$. Since $\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$ and $\dfrac{2\pi}{3}\in[0,\pi]$:

   …

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6. CBSE 2024Set D1 markMCQ
   Q.$x \in [-1, 1],\ \cos^{-1} x =$

   (a) $\frac{\pi}{2} - \cot^{-1} x$

   (b) $\frac{\pi}{2} - \sin^{-1} x$

   (c) $\frac{\pi}{2} - \tan^{-1} x$

   (d) $\frac{\pi}{2} - \sec^{-1} x$

   ›Reveal solutionSolution

   Complementary identity: $\cos^{-1}x=\tfrac\pi2-\sin^{-1}x$ for $x\in[-1,1]$.

   For $x\in[-1,1]$ we have $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$.

   …

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7. CBSE 2024Set D1 markMCQ
   Q.$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) =$

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

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

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

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

   ›Reveal solutionSolution

   $\cos^{-1}$ must return a value in $[0,\pi]$, so the answer is $\tfrac{5\pi}{6}$, not $\tfrac{7\pi}{6}$.

   Since $\tfrac{7\pi}{6}\notin[0,\pi]$, we cannot simply cancel. Compute $\cos\dfrac{7\pi}{6} = -\dfrac{\sqrt3}{2}$.

   …

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8. CBSE 2023Set E1 markMCQ
   Q.$x\in[-1,1],\ \sin\left[2(\sin^{-1}x+\cos^{-1}x)\right] =$

   (a) $0$

   (b) $1$

   (c) $-1$

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

   ›Reveal solutionSolution

   $\sin\left[2(\sin^{-1}x+\cos^{-1}x)\right]=0$.

   For all $x\in[-1,1]$, $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$. Hence

   …

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9. CBSE 2023Set A1 mark
   Q.Find the value of $\cos(\sec^{-1}x + \text{cosec}^{-1}x)$, $|x| \ge 1$.

   ›Reveal solutionSolution

   $\sec^{-1}x+\text{cosec}^{-1}x=\dfrac{\pi}{2}$ for $|x|\ge1$, so $\cos\left(\sec^{-1}x+\text{cosec}^{-1}x\right)=\cos\dfrac{\pi}{2}=0$.

   …

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10. CBSE 2022Set HE2191 mark
    Q.Fill in the blank: $\sin^{-1} x + \cos^{-1} x =$ ______.

    ›Reveal solutionSolution

    This is a standard inverse trigonometric identity valid for $x\in[-1,1]$.

    Let $\sin^{-1}x = \theta$, so $x = \sin\theta = \cos\left(\dfrac{\pi}{2}-\theta\right)$, giving $\cos^{-1}x = \dfrac{\pi}{2}-\theta$. Adding, …

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11. CBSE 2021Set I1 markMCQ
    Q.$\cos^{-1}(2x) + \sin^{-1}(2x) = \ldots\ldots , \ 2x \in [-1,1]$

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

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

    (c) $\pi$

    (d) $0$

    ›Reveal solutionSolution

    For all valid arguments, $\sin^{-1}t+\cos^{-1}t=\tfrac{\pi}{2}$.

    There is a standard identity: for every $t\in[-1,1]$,

    $\sin^{-1}t+\cos^{-1}t=\dfrac{\pi}{2}$.

    …

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12. CBSE 2019Set 65/3/11 mark
    Q.If $y = \sin^{-1} x + \cos^{-1} x$, find $\dfrac{dy}{dx}$.

    ›Reveal solutionSolution

    The sum $\sin^{-1} x + \cos^{-1} x$ is constant ($\pi/2$) for all $x$ in $[-1, 1]$, so its derivative is zero: $\frac{dy}{dx} = 0$.

    Concept and Intuition

    The problem asks for the derivative of $y = \sin^{-1} x + \cos^{-1} x$. A brute-force approach would differentiate each inverse trig function separately using known formulas, then add. But that misses the deeper point.

    There is a beautiful identity: for any $x$ in $[-1, 1]$,

    $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}.$

    Why? Think geometrically. If $\theta = \sin^{-1} x$, then $\sin \theta = x$ and $\theta \in [-\pi/2, \pi/2]$. The complementary angle $\pi/2 - \theta$ has cosine equal to $x$, so $\cos^{-1} x = \pi/2 - \theta$. Adding them gives $\pi/2$.

    Since $y$ is constant, its derivative is zero — no calculation needed. This is the elegant, concept-first way.

    Watch out

    A common mistake is to differentiate each term separately and get $\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$, which is correct but misses why the sum is constant. The identity is the real insight.

    Step-by-Step Solution

    1. Recall the fundamental identity For any $x \in [-1, 1]$,

    $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}.$

    This holds because if $\sin^{-1} x = \theta$, then $\cos(\pi/2 - \theta) = \sin \theta = x$, and $\pi/2 - \theta$ lies in $[0, \pi]$, the principal range of $\cos^{-1}$. …

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