Q.If $3\tan^{-1}x+\cot^{-1}x=\pi$, then $x$ equals
(A) $0$
(B) $1$
(C) $-1$
(D) $\frac{1}{2}$

Showing the 12 most recent of 23 on this concept.

1. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.If $3\tan^{-1}x+\cot^{-1}x=\pi$ then $\sin^{-1}x$ is (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$ (E) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Use $\tan^{-1}x+\cot^{-1}x=\tfrac{\pi}{2}$ to reduce the equation and solve for $x=1$.

   Given $3\tan^{-1}x+\cot^{-1}x=\pi$. Split off the identity term:

   $3\tan^{-1}x+\cot^{-1}x=2\tan^{-1}x+\big(\tan^{-1}x+\cot^{-1}x\big)=2\tan^{-1}x+\frac{\pi}{2}.$ …

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2. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.If $\tan^{-1}x = \tan^{-1}(3) - \dfrac{\pi}{4}$, then $x$ is equal to (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{4}$ (C) $1$ (D) $3$ (E) $2$

   ›Reveal solutionSolution

   Write $\frac{\pi}{4}=\tan^{-1}1$ and use the subtraction formula: $\tan^{-1}3-\tan^{-1}1=\tan^{-1}\frac12\Rightarrow x=\frac12$.

   Apply the identity. $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ with $a=3,b=1$: …

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3. KEAM 2025Set eng-2025-04294 marksMCQ
   Q.If $\tan^{-1}2x + \tan^{-1}3x = \dfrac{\pi}{4}$, then the value of $x$ is equal to (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{2}$ (E) $1$

   ›Reveal solutionSolution

   $\tan^{-1}2x+\tan^{-1}3x=\frac\pi4$ gives $\frac{5x}{1-6x^2}=1$, i.e. $6x^2+5x-1=0$; roots $x=\frac16,-1$, and only $x=\frac16$ satisfies the equation. …

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4. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The value of $\tan^{-1}\left(\frac{\cos x - \sqrt{3}\sin x}{\sqrt{3}\cos x + \sin x}\right)$, where $0 < x < \frac{\pi}{2}$ is (A) $\frac{\pi}{6} - x$ (B) $\frac{\pi}{4} - x$ (C) $\frac{\pi}{3} - x$ (D) $\frac{\pi}{2} - x$ (E) $\pi - x$

   ›Reveal solutionSolution

   Recognize numerator and denominator as $2\cos$ and $2\sin$ of $\left(x+\frac{\pi}{3}\right)$.

   Numerator: $\cos x-\sqrt3\sin x=2\left(\tfrac12\cos x-\tfrac{\sqrt3}{2}\sin x\right)=2\cos\!\left(x+\tfrac{\pi}{3}\right)$.

   Denominator: $\sqrt3\cos x+\sin x=2\left(\tfrac{\sqrt3}{2}\cos x+\tfrac12\sin x\right)=2\sin\!\left(x+\tfrac{\pi}{3}\right)$. …

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5. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.Let $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\sin x - \cos x}\right)$, $0 < x < \frac{\pi}{2}$. Then $\frac{dy}{dx}$ is equal to (A) $x$ (B) $-1$ (C) $-x$ (D) $2x$ (E) $-2x$

   ›Reveal solutionSolution

   Rewrite the fraction using $\sin x\pm\cos x=\sqrt2\,(\ldots)$ to collapse the arctan into a linear function.

   $\sin x+\cos x=\sqrt{2}\cos\left(x-\dfrac{\pi}{4}\right)$ and $\sin x-\cos x=\sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)$.

   So the ratio $=\cot\left(x-\dfrac{\pi}{4}\right)=\tan\left(\dfrac{\pi}{2}-\left(x-\dfrac{\pi}{4}\right)\right)=\tan\left(\dfrac{3\pi}{4}-x\right)$. …

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6. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$cot^{-1}(1) + cot^{-1}(2) + cot^{-1}(3) =$ (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $\frac{3\pi}{2}$ (D) $\pi$ (E) $0$

   ›Reveal solutionSolution

   Convert to $\tan^{-1}$ and use the addition formula.

   $\cot^{-1}1 = \tfrac{\pi}{4}$. For the other two, $\cot^{-1}2 + \cot^{-1}3 = \tan^{-1}\tfrac12 + \tan^{-1}\tfrac13$: …

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7. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.If $y = \tan^{-1}\left[\dfrac{\cos x - \sin x}{\cos x + \sin x}\right]$, $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$, then $\dfrac{dy}{dx}$ is equal to (A) $\tan x$ (B) $\cos x$ (C) $\sin x$ (D) $-1$ (E) $0$

   ›Reveal solutionSolution

   The argument simplifies to $\tan(\frac{\pi}{4}-x)$, so $y=\frac{\pi}{4}-x$ and $\frac{dy}{dx}=-1$.

   Divide numerator and denominator by $\cos x$:

   $y=\tan^{-1}\!\left(\frac{1-\tan x}{1+\tan x}\right)=\tan^{-1}\!\left(\tan\!\left(\frac{\pi}{4}-x\right)\right).$ …

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8. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.If $x\ne 0,\ y\ne 0$, then the value of $\cot^{-1}\left(\dfrac{x}{y}\right) + \cot^{-1}\left(\dfrac{y}{x}\right)$ is (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) 0 (D) $-\pi$ (E) $-\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   $\cot^{-1}t+\cot^{-1}\frac1t=\frac{\pi}{2}$ (for $t>0$).

   Let $t=\frac xy$. Then $\cot^{-1}t=\tan^{-1}\frac1t$ and $\cot^{-1}\frac1t=\tan^{-1}t$, so …

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9. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\tan\left(\tan^{-1}(3)+\tan^{-1}(7)\right)$ is equal to (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{5}$ (D) $-\dfrac{1}{5}$ (E) $0$

   ›Reveal solutionSolution

   Use $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{10}{-20}=-\tfrac12$.

   With $\tan A=3$, $\tan B=7$: …

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10. KEAM 2025Set eng-2025-04274 marksMCQ
    Q.$\tan^{-1}\left(\dfrac{1001}{999}\right)-\tan^{-1}\left(\dfrac{2}{2000}\right) =$ (A) $\dfrac{\pi}{3}$ (B) $\pi$ (C) $1$ (D) $\dfrac{\pi}{6}$ (E) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    Rewrite $\dfrac{2}{2000}=\dfrac{1}{1000}$ and use the arctangent subtraction formula; the numerator and denominator turn out equal, giving $\tan^{-1}(1)=\dfrac{\pi}{4}$.

    We evaluate $\tan^{-1}\!\left(\dfrac{1001}{999}\right)-\tan^{-1}\!\left(\dfrac{2}{2000}\right)$.

    Note $\dfrac{2}{2000}=\dfrac{1}{1000}$. Using $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}$ with $a=\dfrac{1001}{999}$, $b=\dfrac{1}{1000}$: …

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11. KEAM 2021Set eng-2021-P2-B14 marksMCQ
    Q.The value of $\tan^{-1}\left(\dfrac{7}{4}\right) - \tan^{-1}\left(\dfrac{3}{11}\right)$ is equal to (A) $\dfrac{-\pi}{3}$ (B) $\dfrac{-\pi}{4}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{3}$ (E) $\pi$

    ›Reveal solutionSolution

    $\tan^{-1}\tfrac74-\tan^{-1}\tfrac3{11}=\pi/4$.

    Concept and Intuition

    Apply $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$; both arctangents are positive and small enough that the difference lies in the principal range.

    Step-by-Step Solution

    1. Numerator: $\tfrac74-\tfrac3{11}=\tfrac{77-12}{44}=\tfrac{65}{44}$.
    2. Denominator: $1+\tfrac74\cdot\tfrac3{11}=1+\tfrac{21}{44}=\tfrac{65}{44}$. …

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12. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.$\tan^{-1}\left(\dfrac{1}{3}\right)+\tan^{-1}\left(\dfrac{2}{3}\right)+\cot^{-1}\left(\dfrac{9}{7}\right)=$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$ (E) $0$

    ›Reveal solutionSolution

    $\tan^{-1}\tfrac13+\tan^{-1}\tfrac23=\tan^{-1}\dfrac{1/3+2/3}{1-(1/3)(2/3)}=\tan^{-1}\dfrac{1}{7/9}=\tan^{-1}\tfrac97$. Since $\cot^{-1}\tfrac97=\tan^{-1}\tfrac79$, the total is $\tan^{-1}\tfrac97+\tan^{-1}\tfrac79=\dfrac{\pi}{2}$.

    First combine the two arctangents (product $\tfrac13\cdot\tfrac23=\tfrac29<1$, so no correction term):

    $\tan^{-1}\!\tfrac13+\tan^{-1}\!\tfrac23=\tan^{-1}\!\frac{\tfrac13+\tfrac23}{1-\tfrac29}=\tan^{-1}\!\frac{1}{7/9}=\tan^{-1}\!\tfrac97.$ …

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