Q.If $\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}$, then $\cot^{-1}x+\cot^{-1}y$ equals
(A) $\frac{\pi}{5}$
(B) $\frac{2\pi}{5}$
(C) $\frac{3\pi}{5}$
(D) $\pi$

Showing the 12 most recent of 23 on this concept.

1. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.The value of $\tan^{-1}\left(\dfrac{1}{2}\right) + \tan^{-1}\left(\dfrac{2}{5}\right)$ is (A) $\tan^{-1}(5)$ (B) $\tan^{-1}\left(\dfrac{1}{5}\right)$ (C) $\tan^{-1}\left(\dfrac{2}{3}\right)$ (D) $\tan^{-1}\left(\dfrac{8}{9}\right)$ (E) $\tan^{-1}\left(\dfrac{9}{8}\right)$

   ›Reveal solutionSolution

   tan^-1(1/2) + tan^-1(2/5) = tan^-1(9/8) since the product (1/2)(2/5) = 1/5 < 1.

   Concept and Intuition

   tan^-1 a + tan^-1 b = tan^-1((a+b)/(1-ab)) when ab < 1, so the sum stays a single arctangent.

   Step-by-Step Solution

   1. a = 1/2, b = 2/5; ab = 1/5 < 1, so the direct formula applies.
   2. a + b = 1/2 + 2/5 = 9/10.
   3. 1 - ab = 1 - 1/5 = 4/5.
   4. (a+b)/(1-ab) = (9/10)/(4/5) = (9/10)(5/4) = 9/8. …

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2. 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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3. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If $2\cot^{-1}\left(\dfrac{4}{3}\right) = \cos^{-1}\left(\dfrac{x}{5}\right)$ , then the value of $x$ is equal to (A) $\dfrac{3}{25}$ (B) $\dfrac{7}{25}$ (C) $\dfrac{3}{5}$ (D) $\dfrac{7}{5}$ (E) $\dfrac{5}{7}$

   ›Reveal solutionSolution

   Compute $\cos(2\cot^{-1}\tfrac43)=\tfrac{7}{25}$, so $x/5=7/25$ gives $x=\tfrac75$.

   Let $\phi=\cot^{-1}\dfrac43$, so $\cot\phi=\dfrac43$, giving a $3\text{-}4\text{-}5$ triangle with $\sin\phi=\dfrac35$, $\cos\phi=\dfrac45$. Then

   $\cos2\phi=1-2\sin^2\phi=1-2\cdot\frac{9}{25}=\frac{7}{25}.$ …

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4. 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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5. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.$\tan\left(2\tan^{-1}\left(\dfrac{2}{5}\right)\right)$ is equal to (A) $\dfrac{8}{5}$ (B) $\dfrac{10}{21}$ (C) $\dfrac{20}{21}$ (D) $\dfrac{21}{25}$ (E) $\dfrac{4}{25}$

   ›Reveal solutionSolution

   $\tan(2\tan^{-1}\tfrac25)=\tfrac{20}{21}$.

   Concept and Intuition

   The double-angle tangent formula $\tan 2A=\dfrac{2\tan A}{1-\tan^2 A}$ applies with $\tan A=\tfrac25$.

   Step-by-Step Solution

   1. $\tan A=\tfrac25$, so $2\tan A=\tfrac45$ and $\tan^2 A=\tfrac4{25}$.
   2. $1-\tfrac4{25}=\tfrac{21}{25}$. …

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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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