Q.Evaluate $\cos\left[\cos^{-1}\left(\frac{-\sqrt3}{2}\right)+\frac{\pi}{6}\right]$.

Showing the 12 most recent of 66 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.If $2\cos^{-1}x = y$, then (A) $0 \le y \le \pi$ (B) $-\pi \le y \le \pi$ (C) $0 \le y \le 2\pi$ (D) $-\pi \le y \le 0$

   ›Reveal solutionSolution

   The range of $\cos^{-1}x$ is $[0,\pi]$, so multiplying by 2 gives $y = 2\cos^{-1}x$ a range of $[0,2\pi]$. The correct option is (C).

   Concept and Intuition

   The key to this problem lies entirely in understanding the range of the inverse cosine function. $\cos^{-1}x$ (also written as $\arccos x$) is defined as the angle whose cosine is $x$, and by convention, that angle is always taken from the interval $[0, \pi]$. This is not arbitrary — it's the standard principal value branch that makes the function one-to-one and therefore invertible.

   Once you know that $\cos^{-1}x$ lives between $0$ and $\pi$ (inclusive), finding the range of $y = 2\cos^{-1}x$ is simply a matter of scaling that interval by a factor of 2. No tricky domain restrictions, no sign flips — just multiplication.

   Watch out

   A common mistake is to confuse the range of $\cos^{-1}x$ with that of $\sin^{-1}x$ (which is $[-\pi/2, \pi/2]$). Always recall: $\cos^{-1}x \in [0,\pi]$, not $[-\pi/2,\pi/2]$.

   Step-by-step solution

   1. Recall the range of $\cos^{-1}x$ The inverse cosine function $\cos^{-1}: [-1,1] \to [0,\pi]$ gives an output angle in radians. This means:

   $0 \le \cos^{-1}x \le \pi \quad \text{for all } x \in [-1,1].$

   2. Multiply the inequality by 2 Since 2 is positive, multiplying through preserves the direction of the inequalities:

   $2 \cdot 0 \le 2\cos^{-1}x \le 2 \cdot \pi$

   which simplifies to:

   $0 \le y \le 2\pi.$

   3. Check if every value in $[0,2\pi]$ is actually attained

      As $x$ varies continuously from $-1$ to $1$, $\cos^{-1}x$ varies continuously from $\pi$ down to $0$. So $y = 2\cos^{-1}x$ varies continuously from $2\pi$ down to $0$, covering every number in between. The range is exactly the closed interval $[0,2\pi]$.

   4. Match with the given options

      • (A) $0 \le y \le \pi$ — too narrow, misses values between $\pi$ and $2\pi$.
      • (B) $-\pi \le y \le \pi$ — includes negative values, which are impossible since $\cos^{-1}x \ge 0$.
      • (C) $0 \le y \le 2\pi$ — exactly matches our result.
      • (D) $-\pi \le y \le 0$ — entirely negative, completely wrong.

   Tip

   You can also think geometrically: $\cos^{-1}x$ is the angle in the upper half of the unit circle (from $0$ to $\pi$ radians). Doubling that angle sweeps out the full circle's worth of angles — from $0$ all the way around to $2\pi$.

   ✓Final answer

   The correct option is (C), since $y = 2\cos^{-1}x$ lies in $[0, 2\pi]$.

2. CBSE 2026Set V11 markMCQ
   Q.The domain of $\tan^{-1} x$ is

   (a) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$

   (b) $(0, \pi)$

   (c) $[-1, 1]$

   (d) $(-\infty, \infty)$

   ›Reveal solutionSolution

   The tangent function maps $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ onto all of $\mathbb{R}$, so $\tan^{-1}x$ accepts every real $x$; answer (d).

   The principal-branch tangent $\tan:\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\to\mathbb{R}$ is a bijection onto $\mathbb{R}$. Its inverse $\tan^{-1}$ therefore has domain equal to the range of $\tan$, namely all real numbers.

   $\text{Domain}(\tan^{-1}x)=(-\infty,\infty),\qquad \text{Range}=\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right).$

   ✓Final answer

   (d) $(-\infty,\infty)$

3. CBSE 2026Set CX1 mark
   Q.Find the value of $\tan^{-1}\sqrt{3}-\sec^{-1}(-2)$.

   ›Reveal solutionSolution

   $\tan^{-1}\sqrt3=\tfrac{\pi}{3}$, $\sec^{-1}(-2)=\tfrac{2\pi}{3}$, giving $-\tfrac{\pi}{3}$.

   Concept: Use the principal-value ranges: $\tan^{-1}\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ and $\sec^{-1}\in[0,\pi]\setminus\{\tfrac{\pi}{2}\}$.

   $\tan^{-1}\sqrt3=\dfrac{\pi}{3}\qquad(\tan\tfrac{\pi}{3}=\sqrt3).$

   For $\sec^{-1}(-2)$ we need $\theta\in[0,\pi]$ with $\sec\theta=-2$, i.e. $\cos\theta=-\tfrac12$, giving $\theta=\dfrac{2\pi}{3}$.

   $\tan^{-1}\sqrt3-\sec^{-1}(-2)=\dfrac{\pi}{3}-\dfrac{2\pi}{3}=-\dfrac{\pi}{3}.$

   ✓Final answer

   $\tan^{-1}\sqrt3-\sec^{-1}(-2)=-\dfrac{\pi}{3}$.

4. CBSE 2026Set ANNUAL1 mark
   Q.$\sin^{-1}x$ is a function whose domain is __________.

   ›Reveal solutionSolution

   $\sin^{-1}x$ is defined only where $\sin\theta=x$ has a solution, i.e. for $x\in[-1,1]$.

   The sine function takes values only in $[-1,1]$, so its inverse $\sin^{-1}x$ can only accept inputs in that range.

   ✓Final answer

   The domain of $\sin^{-1}x$ is $[-1,1]$.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $y = \cos^{-1} x$ then

   (a) $0 \le y \le \pi$

   (b) $-\dfrac{\pi}{2} \le y \le \dfrac{\pi}{2}$

   (c) $-\pi \le y \le \pi$

   (d) None of these

   ›Reveal solutionSolution

   $\cos^{-1}x$ is defined so that its principal value always lies in $[0,\pi]$.

   The function $\cos x$ is one-one and onto from $[0,\pi]$ to $[-1,1]$, so its inverse $\cos^{-1}x$ is defined on domain $[-1,1]$ with range (principal value branch) $[0,\pi]$. Thus if $y=\cos^{-1}x$, then $0 \le y \le \pi$.

   ✓Final answer

   (a) $0 \le y \le \pi$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Principal value of $\tan^{-1}(-1)$ is

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

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

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

   (d) None of these

   ›Reveal solutionSolution

   The principal value of $\tan^{-1}x$ always lies in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

   We need $y$ such that $\tan y = -1$ and $y \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$.

   Since $\tan\left(\dfrac{\pi}{4}\right) = 1$, we get $\tan\left(-\dfrac{\pi}{4}\right) = -1$, and $-\dfrac{\pi}{4}$ lies within the principal branch.

   So $\tan^{-1}(-1) = -\dfrac{\pi}{4}$.

   ✓Final answer

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

7. CBSE 2026Set ANNUAL1 markMCQ
   Q.The principal value of $\cos^{-1} x$ is:

   (a) $[0, \pi]$

   (b) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

   (c) $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

   (d) None of these

   ›Reveal solutionSolution

   The principal value branch of $\cos^{-1}x$ is $[0,\pi]$ by definition.

   The function $\cos:[0,\pi]\to[-1,1]$ is a bijection, so its inverse $\cos^{-1}:[-1,1]\to[0,\pi]$ is defined with range (principal value branch) $[0,\pi]$.

   ✓Final answer

   Option (a): $[0,\pi]$.

8. CBSE 2026Set ANNUAL1 markMCQ
   Q.Principal value of cos⁻¹(1/2) is:

   (a) π/2

   (b) π/3

   (c) π/4

   (d) π/6

   ›Reveal solutionSolution

   The principal value of $\cos^{-1}x$ lies in $[0,\pi]$, and $\cos\left(\frac{\pi}{3}\right) = \frac12$.

   We need $\theta \in [0,\pi]$ such that $\cos\theta = \dfrac12$.

   Since $\cos\left(\dfrac{\pi}{3}\right) = \dfrac12$ and $\dfrac{\pi}{3} \in [0,\pi]$, this is the principal value.

   ✓Final answer

   $\cos^{-1}\left(\dfrac12\right) = \dfrac{\pi}{3}$ (option b).

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

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

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

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

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

   ›Reveal solutionSolution

   Find the angle in the principal-value range $[0,\pi]$ of $\cos^{-1}$ whose cosine equals $\dfrac{\sqrt3}{2}$.

   We need $\theta\in[0,\pi]$ (the principal-value branch of $\cos^{-1}$) such that

   $\cos\theta = \frac{\sqrt3}{2}$

   From the standard trigonometric ratio table, $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$, and $\dfrac{\pi}{6}$ lies in $[0,\pi]$.

   $\cos^{-1}\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{6}$

   ✓Final answer

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

10. CBSE 2026Set ANNUAL1 markMCQ
    Q.The principal value of cot⁻¹(-1/√3) is ................. .

    (a) π/3

    (b) π/4

    (c) 2π/3

    (d) 4π/3

    ›Reveal solutionSolution

    The principal range of $\cot^{-1}$ is $(0,\pi)$; find the angle in that range with cotangent $-1/\sqrt3$.

    We know $\cot(\pi/3)=1/\sqrt3$. Since the given value is negative and the principal range of $\cot^{-1}$ is $(0,\pi)$, the required angle lies in the second quadrant, where cotangent is negative.

    $\cot(2\pi/3) = \dfrac{\cos(2\pi/3)}{\sin(2\pi/3)} = \dfrac{-1/2}{\sqrt3/2} = -\dfrac{1}{\sqrt3}$

    Since $2\pi/3 \in (0,\pi)$, this is the principal value.

    ✓Final answer

    $\cot^{-1}(-1/\sqrt3) = 2\pi/3$ — option (c).

11. CBSE 2026Set ANNUAL1 markMCQ
    Q.The principal value of tan^{-1}(-\sqrt{3}) is:

    (a)

    (i) \pi/3

    (b)

    (ii) -\pi/3

    (c)

    (iii) \pi/6

    (d)

    (iv) -\pi/6

    ›Reveal solutionSolution

    $\tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}$ — option (ii).

    Concept. The principal value of $\tan^{-1} x$ is the unique angle $\theta$ lying in the open interval $\left(-\frac{\pi}{2},\ \frac{\pi}{2}\right)$ such that $\tan\theta = x$. This is the standard NCERT/CBSE principal-value branch that the UBSE Class-12 syllabus also follows.

    Why this branch. Tangent is one-to-one on $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so exactly one angle there gives each real value.

    Steps.

    • We need $\theta$ with $\tan\theta = -\sqrt{3}$ and $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$.

    • $\tan\dfrac{\pi}{3} = \sqrt{3}$, and $\tan$ is an odd function, so $\tan\left(-\dfrac{\pi}{3}\right) = -\sqrt{3}$.

    • $-\dfrac{\pi}{3}$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, so it is the principal value.

    ✓Final answer

    $\tan^{-1}(-\sqrt{3}) = -\dfrac{\pi}{3}$ — option (ii).

12. CBSE 2025Set X11 markMCQ
    Q.Match List - I with List - II.

    List - I	List - II
    A) Domain of $\sin^{-1} x$	i) $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$
    B) Range of $\tan^{-1} x$	ii) $[0, \pi]$
    C) Range of $\cos^{-1} x$	iii) $[-1, 1]$

    Choose the correct answer from the options given below :

    (a) A-i, B-ii, C-iii

    (b) A-iii, B-ii, C-i

    (c) A-ii, B-i, C-iii

    (d) A-iii, B-i, C-ii

    ›Reveal solutionSolution

    Matching inverse-trig domains/ranges — correct option is (d).

    Recall the standard facts: the domain of $\sin^{-1}x$ is $[-1,1]$ (matches iii), the principal range of $\tan^{-1}x$ is the open interval $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ (matches i), and the principal range of $\cos^{-1}x$ is the closed interval $[0,\pi]$ (matches ii). Hence A-iii, B-i, C-ii.

    ✓Final answer

    (d) A-iii, B-i, C-ii