Q.Find $\frac{dy}{dx}$ in the following: $y = \tan^{-1} \left(\frac{3x-x^3}{1-3x^2}\right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$

Showing the 12 most recent of 23 on this concept.

1. 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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2. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.If $f(x) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $f\left(\dfrac{1}{\sqrt{3}}\right)$ is equal to (A) $\dfrac{\pi}{6}$ (B) $\dfrac{2\pi}{3}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{4\pi}{3}$ (E) $0$

   ›Reveal solutionSolution

   Using tan^-1(2x/(1-x^2)) = 2 tan^-1 x for |x|<1, f(1/sqrt3) = 2*(pi/6) = pi/3.

   Concept and Intuition

   The identity 2 tan^-1 x = tan^-1(2x/(1-x^2)) holds for |x| < 1. Since 1/sqrt3 < 1, the formula applies directly.

   Step-by-Step Solution

   1. f(x) = tan^-1(2x/(1-x^2)) = 2 tan^-1 x for |x| < 1.
   2. tan^-1(1/sqrt3) = pi/6.
   3. f(1/sqrt3) = 2 * pi/6 = pi/3. …

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3. 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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4. 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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5. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.The value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}\left(\dfrac{2}{\sqrt{3}}\right)$ is (A) $\dfrac{2\pi}{3}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$ (E) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   tan^-1(sqrt3) = pi/3 and sec^-1(2/sqrt3) = pi/6, so the difference is pi/6.

   Concept and Intuition

   Both are standard angles: tan(pi/3) = sqrt3, and sec(pi/6) = 1/cos(pi/6) = 2/sqrt3.

   Step-by-Step Solution

   1. tan^-1(sqrt3) = pi/3.
   2. sec^-1(2/sqrt3): cos of the angle = sqrt3/2, so the angle = pi/6. …

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6. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.$\tan^{-1} 2 - \tan^{-1}\left(\frac{1}{3}\right)$ is equal to (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{6}$ (E) $0$

   ›Reveal solutionSolution

   Apply the subtraction formula $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}$.

   With $a=2,\ b=\dfrac13$: …

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7. 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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8. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.If $a=\tan^{-1}\left(\dfrac{4}{3}\right)$ and $b=\tan^{-1}\left(\dfrac{1}{3}\right)$, where $0<a,b<\dfrac{\pi}{2}$, then $a-b=$ (A) $\tan^{-1}(3)$ (B) $\tan^{-1}\left(\dfrac{3}{13}\right)$ (C) $\tan^{-1}(5)$ (D) $\tan^{-1}\left(\dfrac{9}{13}\right)$ (E) $\tan^{-1}\left(\dfrac{5}{13}\right)$

   ›Reveal solutionSolution

   Apply the arctangent subtraction formula.

   With $\tan a=\dfrac{4}{3}$ and $\tan b=\dfrac{1}{3}$:

   $\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}=\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{3}\cdot\frac{1}{3}}=\frac{1}{1+\frac{4}{9}}=\frac{1}{\frac{13}{9}}=\frac{9}{13}.$ …

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9. 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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10. 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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11. 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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12. KEAM 2026Set eng-2026-04214 marksMCQ
    Q.The value of $2\tan^{-1}\left(\frac{1}{3}\right)+\cot^{-1}\left(\frac{3}{4}\right)=$ (A) $\frac{\pi}{3}$ (B) $\frac{2\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{6}$ (E) $\frac{\pi}{2}$

    ›Reveal solutionSolution

    $2\tan^{-1}\frac13=\tan^{-1}\frac34$; adding $\tan^{-1}\frac43$ (its complementary reciprocal) gives $\frac{\pi}{2}$.

    Using the double-angle formula:

    $2\tan^{-1}\frac13=\tan^{-1}\!\frac{2\cdot\frac13}{1-\frac19}=\tan^{-1}\!\frac{\frac23}{\frac89}=\tan^{-1}\frac34.$ …

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