Q.Find the value of $\tan^{-1}\left(\tan\frac{2\pi}{3}\right)$.

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\mathrm{Tan}^{-1} 2 + \mathrm{Tan}^{-1} 3 =$ (A) $-\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{3\pi}{4}$ (D) $\dfrac{5\pi}{4}$

   ›Reveal solutionSolution

   Adding two inverse-tangent principal values whose tangent-sum formula gives $-1$, the actual sum is $3\pi/4$ (not $-\pi/4$), because both individual angles are obtuse-leaning acute angles summing past $\pi/2$.

   Concept and Intuition

   The tangent addition formula only gives $\tan(A+B)$, not $A+B$ directly — since tangent is periodic with period $\pi$, we must use the actual sizes of $A=\tan^{-1}2$ and $B=\tan^{-1}3$ (each in $(0,\pi/2)$, and in fact each $>\pi/4$ since $\tan>1$) to determine which branch the sum falls into.

   Step-by-Step Solution

   1. Let $A=\tan^{-1}2$, $B=\tan^{-1}3$; both lie in $(\pi/4,\pi/2)$ since $\tan A=2>1,\tan B=3>1$.
   2. $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{2+3}{1-6}=\dfrac{5}{-5}=-1$.
   3. Since $A,B\in(\pi/4,\pi/2)$, their sum $A+B\in(\pi/2,\pi)$. …

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2. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.$\text{Tan}^{-1}(-2) - \text{Tan}^{-1}(3)$ is equal to (A) $\dfrac{3\pi}{4}$ (B) $\dfrac{-\pi}{6}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{-3\pi}{4}$

   ›Reveal solutionSolution

   Convert to $-(\tan^{-1}2+\tan^{-1}3)$ and apply the addition formula (with the $+\pi$ correction since $ab>1$) to get $-\frac{3\pi}{4}$.

   Concept and Intuition

   $\tan^{-1}$ is an odd function, so $\tan^{-1}(-x)=-\tan^{-1}(x)$. The standard addition formula $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}$ needs a $+\pi$ correction whenever $a,b>0$ and $ab>1$, because then the true sum exceeds $\pi/2$ while the raw arctan formula would return a negative principal value.

   Step-by-Step Solution

   1. $\tan^{-1}(-2) = -\tan^{-1}(2)$, so the expression becomes $-\tan^{-1}(2)-\tan^{-1}(3)$.
   2. Compute $\tan^{-1}2+\tan^{-1}3$. Here $a=2,b=3$, $ab=6>1$, both positive, so use $\tan^{-1}a+\tan^{-1}b = \pi+\tan^{-1}\frac{a+b}{1-ab}$.
   3. $\frac{a+b}{1-ab} = \frac{5}{1-6}=\frac{5}{-5}=-1$, and $\tan^{-1}(-1)=-\frac{\pi}{4}$. …

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3. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\tan\left(2\,\mathrm{Tan}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Tan}^{-1}\left(\dfrac{1}{7}\right)\right) =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $\sqrt{3}$ (C) $1$ (D) $3/7$

   ›Reveal solutionSolution

   First reduce the double-angle inverse-tangent term to a single tangent value using the tangent double-angle formula, then combine with the second $\tan^{-1}$ term using the tangent addition formula. The result is exactly $1$.

   Concept and Intuition

   Expressions like $\tan(2\tan^{-1}x+\tan^{-1}y)$ are handled in two stages: first collapse $2\tan^{-1}x$ to a single angle whose tangent is known via the double-angle formula $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$, then treat the whole thing as $\tan(\alpha'+\beta)$ using the standard addition formula, where $\alpha'$ is the angle with $\tan\alpha'=\tan(2\tan^{-1}x)$.

   Step-by-Step Solution

   1. Let $\alpha=\tan^{-1}(1/3)$, so $\tan\alpha=1/3$. Then $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac{2}{3}\times\dfrac{9}{8}=\dfrac{18}{24}=\dfrac34$.
   2. Let $\beta=\tan^{-1}(1/7)$, so $\tan\beta=1/7$.
   3. We need $\tan(2\alpha+\beta)=\dfrac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta}=\dfrac{\frac34+\frac17}{1-\frac34\cdot\frac17}$. …

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4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$4\tan^{-1}\dfrac{1}{5} - \tan^{-1}\dfrac{1}{70} + \tan^{-1}\dfrac{1}{99} =$ (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   This is a Machin-like arctangent identity that evaluates exactly to $\pi/4$.

   Concept and Intuition

   Sums/differences of $\tan^{-1}$ terms with small reciprocal arguments often combine (via repeated use of $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\frac{p-q}{1+pq}$ and the double/quadruple-angle formula for tangent) into a single nice angle like $\pi/4$. These are the classical 'Machin-type' formulas historically used to compute $\pi$.

   Step-by-Step Solution

   1. First combine $4\tan^{-1}\frac15$ using the double-angle formula for $\tan$ twice: with $\tan\alpha=\frac15$, $\tan2\alpha=\dfrac{2\cdot\frac15}{1-\frac1{25}}=\dfrac{2/5}{24/25}=\dfrac{5}{12}$, and $\tan4\alpha=\dfrac{2\cdot\frac5{12}}{1-\frac{25}{144}}=\dfrac{5/6}{119/144}=\dfrac{120}{119}$.
   2. So $4\tan^{-1}\frac15 = \tan^{-1}\dfrac{120}{119}$ (in the correct quadrant, since $\frac{120}{119}$ is only slightly bigger than 1, the angle is just over $\pi/4$).
   3. Now combine $\tan^{-1}\dfrac{120}{119}-\tan^{-1}\dfrac1{70}$ using $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\frac{p-q}{1+pq}$: numerator $\frac{120}{119}-\frac1{70} = \frac{120\cdot70-119}{119\cdot70}=\frac{8400-119}{8330}=\frac{8281}{8330}$; denominator $1+\frac{120}{119\cdot70}=1+\frac{120}{8330}=\frac{8450}{8330}$. Ratio $=\frac{8281}{8450}$. …

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5. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.What is the value of $\mathrm{Sin}^{-1}\dfrac{12}{13} + \mathrm{Cos}^{-1}\dfrac{4}{5} + \mathrm{Tan}^{-1}\dfrac{63}{16} =$ (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{3\pi}{4}$

   ›Reveal solutionSolution

   The three inverse-trig terms telescope to exactly $\pi$ because the first two sum to $180^\circ$ minus the third.

   Concept and Intuition

   $\sin^{-1}\frac{12}{13}$ is the acute angle with sine $\frac{12}{13}$, cosine $\frac{5}{13}$, tangent $\frac{12}{5}$. $\cos^{-1}\frac{4}{5}$ is the acute angle with cosine $\frac{4}{5}$, sine $\frac{3}{5}$, tangent $\frac{3}{4}$. Adding two acute angles whose tangent-sum formula gives a negative tangent tells us their sum exceeds $90^\circ$ — a classic trick for handling sums of inverse trig terms without a calculator.

   Step-by-Step Solution

   1. Let $\alpha=\sin^{-1}\frac{12}{13}$ (so $\tan\alpha=\frac{12}{5}$) and $\beta=\cos^{-1}\frac{4}{5}$ (so $\tan\beta=\frac{3}{4}$).
   2. $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}=\dfrac{\frac{63}{20}}{-\frac{16}{20}}=-\dfrac{63}{16}$.
   3. Since $\alpha\approx 67.4^\circ,\beta\approx 36.9^\circ$, their sum $\alpha+\beta\approx 104.3^\circ\in(90^\circ,180^\circ)$, where tangent is negative — consistent. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\mathrm{Tanh}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Coth}^{-1}(3) =$ (A) $\mathrm{Sech}^{-1}\left(\dfrac{1}{3}\right)$ (B) $\mathrm{Cosech}^{-1}\left(\dfrac{1}{3}\right)$ (C) $\mathrm{Cosh}^{-1}\left(\dfrac{4}{3}\right)$ (D) $\mathrm{Sinh}^{-1}\left(\dfrac{3}{4}\right)$

   ›Reveal solutionSolution

   This tests the identity linking $\mathrm{Coth}^{-1}$ to $\mathrm{Tanh}^{-1}$ and the logarithmic form of inverse hyperbolic functions; the sum collapses to $\log 2 = \mathrm{Sinh}^{-1}(3/4)$.

   Concept and Intuition

   Inverse hyperbolic functions all reduce to logarithms. For $|x|>1$, $\mathrm{Coth}^{-1}(x) = \mathrm{Tanh}^{-1}(1/x)$ because $\coth\theta = x \iff \tanh\theta = 1/x$. This lets us rewrite both terms of the sum using the SAME inverse function, so they simply add.

   Step-by-Step Solution

   1. Since $3>1$, use $\mathrm{Coth}^{-1}(3) = \mathrm{Tanh}^{-1}(1/3)$.
   2. The sum becomes $\mathrm{Tanh}^{-1}(1/3) + \mathrm{Tanh}^{-1}(1/3) = 2\,\mathrm{Tanh}^{-1}(1/3)$.
   3. Use $\mathrm{Tanh}^{-1}(y) = \dfrac12 \log\!\left(\dfrac{1+y}{1-y}\right)$ with $y = 1/3$: $\mathrm{Tanh}^{-1}(1/3) = \dfrac12\log\!\left(\dfrac{4/3}{2/3}\right) = \dfrac12 \log 2$.
   4. So the sum $= 2\times \dfrac12\ln2 = \log 2$.
   5. Test $\mathrm{Sinh}^{-1}(3/4) = \log\!\left(y+\sqrt{y^2+1}\right)$ with $y=3/4$: $\log\!\left(\dfrac34+\sqrt{\dfrac{9}{16}+1}\right) = \log\!\left(\dfrac34+\dfrac54\right) = \log 2$. Exact match. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\tan^{-1}\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{15}+1} + \tan^{-1}\dfrac{1}{\sqrt{5}} =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   This tests simplifying a nested surd inside an inverse trig function, then adding two inverse-tangent values. The answer is (A).

   Concept and Intuition

   A nested surd of the form $\sqrt{a+b-2\sqrt{ab}}$ always simplifies to $|\sqrt a-\sqrt b|$. Recognizing $8-2\sqrt{15}$ as this form with $a=5,b=3$ turns an ugly expression into a clean one, after which the sum of the two arctangents can be identified as a standard angle.

   Step-by-Step Solution

   1. Write $8-2\sqrt{15}=5+3-2\sqrt{5\cdot3}=(\sqrt5-\sqrt3)^2$, so $\sqrt{8-2\sqrt{15}}=\sqrt5-\sqrt3$ (positive since $\sqrt5>\sqrt3$).
   2. The first term becomes $\tan^{-1}\dfrac{\sqrt5-\sqrt3}{\sqrt{15}+1}$.
   3. Numerically: $\sqrt5\approx2.236$, $\sqrt3\approx1.732$, $\sqrt{15}\approx3.873$. So the argument $\approx\dfrac{0.504}{4.873}\approx0.1034$, giving the first term $\approx5.91°$. …

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8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\dfrac{1}{2}\sin^{-1}\left(\dfrac{3\sin 2\theta}{5+4\cos 2\theta}\right) = \tan^{-1}x$ then $x =$ (A) $\tan\dfrac{\theta}{3}$ (B) $\dfrac{1}{3}\tan\theta$ (C) $\tan 3\theta$ (D) $\dfrac{1}{3}\tan 3\theta$

   ›Reveal solutionSolution

   Rewrite the inverse-sine argument as $\sin(2\phi)$ for a suitable $\tan\phi$ built from $\tan\theta$, so the half of the inverse sine collapses to $\tan^{-1}$ of that quantity directly. Answer: $x=\dfrac13\tan\theta$.

   Concept and Intuition

   Expressions of the form $\dfrac{k\sin2\theta}{A+B\cos2\theta}$ can often be massaged, after dividing through by $(1+\tan^2\theta)$ i.e. converting to $t=\tan\theta$, into the recognizable double-angle sine pattern $\dfrac{2u}{1+u^2}=\sin(2\tan^{-1}u)$ for some rescaled variable $u$. Spotting this pattern converts an awkward inverse-trig expression into a clean angle.

   Step-by-Step Solution

   1. Let $t=\tan\theta$. Recall $\sin2\theta=\dfrac{2t}{1+t^2}$, $\cos2\theta=\dfrac{1-t^2}{1+t^2}$.
   2. Numerator: $3\sin2\theta=\dfrac{6t}{1+t^2}$.
   3. Denominator: $5+4\cos2\theta=5+\dfrac{4(1-t^2)}{1+t^2}=\dfrac{5(1+t^2)+4(1-t^2)}{1+t^2}=\dfrac{9+t^2}{1+t^2}$.
   4. So the ratio is $\dfrac{6t/(1+t^2)}{(9+t^2)/(1+t^2)}=\dfrac{6t}{9+t^2}$.
   5. Let $u=t/3$. Then $\dfrac{2u}{1+u^2}=\dfrac{2t/3}{1+t^2/9}=\dfrac{2t/3}{(9+t^2)/9}=\dfrac{6t}{9+t^2}$ — exactly matches!
   6. So $\dfrac{6t}{9+t^2}=\sin(2\phi)$ where $\tan\phi=u=t/3$, i.e. $\phi=\tan^{-1}\!\left(\dfrac{\tan\theta}{3}\right)$. …

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9. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\sin\left(2\tan^{-1}\left(\dfrac{1}{3}\right)\right) + \cos\left(\tan^{-1}2\sqrt{2}\right) =$ (A) $\dfrac{16}{15}$ (B) $\dfrac{14}{15}$ (C) $\dfrac{11}{15}$ (D) $\dfrac{8}{15}$

   ›Reveal solutionSolution

   This tests reading sine/cosine of an inverse-tangent angle off a right triangle, then applying the double-angle sine formula. The sum evaluates to $14/15$.

   Concept and Intuition

   Given $\tan^{-1}(p/q)$, build the right triangle with opposite $p$, adjacent $q$, hypotenuse $\sqrt{p^2+q^2}$; then any trig function of that angle is a direct ratio of the triangle's sides. This avoids working with inverse functions directly.

   Step-by-Step Solution

   1. Let $\alpha=\tan^{-1}(1/3)$: right triangle with opposite $1$, adjacent $3$, hypotenuse $\sqrt{1+9}=\sqrt{10}$. So $\sin\alpha=\dfrac{1}{\sqrt{10}}$, $\cos\alpha=\dfrac{3}{\sqrt{10}}$.
   2. $\sin2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\dfrac{1}{\sqrt{10}}\cdot\dfrac{3}{\sqrt{10}} = \dfrac{6}{10}=\dfrac{3}{5}$. …

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10. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.The real values of $x$ that satisfy the equation $\tan^{-1}x + \tan^{-1}2x = \dfrac{\pi}{4}$ is (A) $\dfrac{-3\pm\sqrt{17}}{4}$ (B) $-1\pm\sqrt{3}$ (C) $\sqrt{3}-1$ (D) $\dfrac{\sqrt{17}-3}{4}$

    ›Reveal solutionSolution

    The key idea is to apply the inverse tangent addition formula $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab}$ (with a domain check) and then solve the resulting quadratic, finally verifying which solutions satisfy the original equation. The only valid solution is $\frac{\sqrt{17}-3}{4}$, which corresponds to option (D).

    ---

    We start with the equation

    $\tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4}.$

    1. Recall the inverse tangent addition formula

    For real numbers $a$ and $b$ with $ab \neq 1$, we have

    $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab} + k\pi,$

    where $k$ is an integer chosen so that the sum lies in $(-\pi/2, \pi/2)$ (the principal range of $\arctan$).

    Since the right-hand side is $\pi/4$, which is within $(-\pi/2, \pi/2)$, we can safely take $k=0$ provided the sum of the two angles is indeed in that interval. We’ll check this later.

    2. Apply the formula

    Set $a = x$, $b = 2x$. Then

    $\tan^{-1}x + \tan^{-1}2x = \tan^{-1}\frac{x + 2x}{1 - x \cdot 2x} = \tan^{-1}\frac{3x}{1 - 2x^2}.$

    Thus the equation becomes

    $\tan^{-1}\frac{3x}{1 - 2x^2} = \frac{\pi}{4}.$

    3. Remove the arctangent

    Taking tangent of both sides (valid because both sides lie in $(-\pi/2, \pi/2)$ for the moment), we get

    $\frac{3x}{1 - 2x^2} = \tan\frac{\pi}{4} = 1.$

    4. Solve the resulting equation

    $\frac{3x}{1 - 2x^2} = 1 \quad\Rightarrow\quad 3x = 1 - 2x^2.$

    Rearrange:

    $2x^2 + 3x - 1 = 0.$

    Solve using the quadratic formula:

    $x = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}.$

    So the two candidates are

    $x_1 = \frac{-3 + \sqrt{17}}{4}, \qquad x_2 = \frac{-3 - \sqrt{17}}{4}.$

    5. Check domain and validity

    We must ensure that for each candidate, the original sum of arctangents actually equals $\pi/4$ (not $\pi/4 + \pi$ or something else).

    • For $x_2 = \frac{-3 - \sqrt{17}}{4}$: …

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11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.$\dfrac{d}{dx}\left(\operatorname{Tan}^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)\right)=$ (A) $\dfrac{1}{2}$ (B) $\dfrac{-1}{2}$ (C) $1$ (D) $-1$

    ›Reveal solutionSolution

    Simplify the argument to a single tangent of a half-angle expression, then differentiate. Answer: $-\dfrac12$.

    Concept and Intuition

    Expressions like $\dfrac{\cos x}{1+\sin x}$ are classic half-angle simplifications that collapse to $\tan\left(\dfrac\pi4-\dfrac x2\right)$, letting the inverse tangent cancel with the tangent directly.

    Step-by-Step Solution

    1. Multiply numerator and denominator by $(1-\sin x)$: $\dfrac{\cos x}{1+\sin x}=\dfrac{\cos x(1-\sin x)}{1-\sin^2x}=\dfrac{\cos x(1-\sin x)}{\cos^2x}=\dfrac{1-\sin x}{\cos x}$.
    2. This is a known identity: $\dfrac{1-\sin x}{\cos x}=\tan\left(\dfrac\pi4-\dfrac x2\right)$. …

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12. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.$\operatorname{Tan}^{-1}\left(\dfrac{1}{2}\right) + \operatorname{Tan}^{-1}\left(\dfrac{1}{8}\right) + \operatorname{Tan}^{-1}\left(\dfrac{1}{18}\right) + \operatorname{Tan}^{-1}\left(\dfrac{1}{32}\right) =$ (A) $\operatorname{Tan}^{-1}\left(\dfrac{3}{5}\right)$ (B) $\operatorname{Tan}^{-1}\left(\dfrac{5}{8}\right)$ (C) $\operatorname{Tan}^{-1}\left(\dfrac{3}{4}\right)$ (D) $\operatorname{Tan}^{-1}\left(\dfrac{4}{5}\right)$

    ›Reveal solutionSolution

    Recognize the telescoping identity $\tan^{-1}\frac{1}{2n^2}=\tan^{-1}\frac{1}{2n-1}-\tan^{-1}\frac{1}{2n+1}$; the four given terms ($n=1,2,3,4$) telescope down to $\tan^{-1}1-\tan^{-1}\frac19=\tan^{-1}\frac45$.

    Concept and Intuition

    The denominators $2,8,18,32$ are exactly $2\cdot1^2,\ 2\cdot2^2,\ 2\cdot3^2,\ 2\cdot4^2$ — a strong hint to use the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}$ in reverse: for consecutive odd-reciprocal terms $\frac{1}{2n-1}$ and $\frac{1}{2n+1}$, their difference is exactly $\tan^{-1}\frac{1}{2n^2}$. This turns the whole sum into a telescoping series where all the intermediate terms cancel.

    Step-by-Step Solution

    1. Verify the identity for general $n$: $\dfrac{\frac{1}{2n-1}-\frac{1}{2n+1}}{1+\frac{1}{(2n-1)(2n+1)}} = \dfrac{\frac{2}{4n^2-1}}{\frac{4n^2-1+1}{4n^2-1}} = \dfrac{2}{4n^2}=\dfrac{1}{2n^2}$. So $\tan^{-1}\frac{1}{2n^2}=\tan^{-1}\frac{1}{2n-1}-\tan^{-1}\frac{1}{2n+1}$.
    2. Apply with $n=1,2,3,4$:

    • $\tan^{-1}\frac12 = \tan^{-1}1-\tan^{-1}\frac13$
    • $\tan^{-1}\frac18 = \tan^{-1}\frac13-\tan^{-1}\frac15$
    • $\tan^{-1}\frac{1}{18} = \tan^{-1}\frac15-\tan^{-1}\frac17$
    • $\tan^{-1}\frac{1}{32} = \tan^{-1}\frac17-\tan^{-1}\frac19$

    3. Summing all four, the intermediate terms $\tan^{-1}\frac13,\tan^{-1}\frac15,\tan^{-1}\frac17$ cancel in pairs (telescoping), leaving: …

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