Q.Show that $2\tan^{-1}(-3)=\frac{-\pi}{2}+\tan^{-1}\left(\frac{-4}{3}\right)$.

Showing the 12 most recent of 23 on this concept.

1. 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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2. 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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3. 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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4. 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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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 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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7. 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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8. 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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9. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.The value of $\tan\left[\tan^{-1}\left(\dfrac{3}{4}\right) + \tan^{-1}\left(\dfrac{2}{3}\right)\right]$ is (A) $\dfrac{17}{6}$ (B) $\dfrac{6}{17}$ (C) $\dfrac{-17}{6}$ (D) $\dfrac{-6}{11}$ (E) $1$

   ›Reveal solutionSolution

   Apply the tangent addition formula directly to the two arctangents.

   Let $A=\tan^{-1}\left(\dfrac{3}{4}\right)$ and $B=\tan^{-1}\left(\dfrac{2}{3}\right)$, so $\tan A=\dfrac{3}{4}$ and $\tan B=\dfrac{2}{3}$.

   $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{\dfrac{3}{4}+\dfrac{2}{3}}{1-\dfrac{3}{4}\cdot\dfrac{2}{3}}$ …

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10. 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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11. 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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12. 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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