Q.State True or False: The minimum value of $n$ for which $\tan^{-1}\frac{n}{\pi}>\frac{\pi}{4}$, $n\in N$, is valid is 5.

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 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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3. 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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4. 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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5. 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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6. 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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7. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $\theta = 2\tan^{-1}\dfrac{1}{8} + 2\tan^{-1}\dfrac{1}{5} + \tan^{-1}\dfrac{1}{7}$ and $\tan\dfrac{\theta}{2} = \sqrt{m} + \sqrt{n}$ where $m$ and $n$ are positive integers such that $m < n$ then $(m^n + n^m)^{m+n} =$ (A) $18$ (B) $27$ (C) $25$ (D) $36$

   ›Reveal solutionSolution

   Simplifying the sum of inverse tangents gives $\theta=\pi/4$, so $\tan(\theta/2)=\tan(\pi/8)=\sqrt2-1$; with $m=1,n=2$ this gives $(m^n+n^m)^{m+n}=27$.

   Concept and Intuition

   Repeated use of the tangent addition formula $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ collapses the sum of arctangents into a single, simple angle.

   Step-by-Step Solution

   1. $\tan^{-1}\dfrac18+\tan^{-1}\dfrac15 = \tan^{-1}\!\left(\dfrac{1/8+1/5}{1-1/40}\right)=\tan^{-1}\!\left(\dfrac{13/40}{39/40}\right)=\tan^{-1}\dfrac13$.
   2. So $2\tan^{-1}\dfrac18+2\tan^{-1}\dfrac15 = 2\tan^{-1}\dfrac13$.
   3. Double-angle: $\tan\left(2\tan^{-1}\tfrac13\right)=\dfrac{2\cdot\frac13}{1-\frac19}=\dfrac{2/3}{8/9}=\dfrac34$, so $2\tan^{-1}\tfrac13=\tan^{-1}\tfrac34$.
   4. $\theta=\tan^{-1}\dfrac34+\tan^{-1}\dfrac17 = \tan^{-1}\!\left(\dfrac{3/4+1/7}{1-3/28}\right)=\tan^{-1}\!\left(\dfrac{25/28}{25/28}\right)=\tan^{-1}(1)=\dfrac{\pi}{4}$. …

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8. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $\tan^{-1}\left[\dfrac{1}{1+1.2}\right] + \tan^{-1}\left[\dfrac{1}{1+2.3}\right] + \cdots + \tan^{-1}\left[\dfrac{1}{1+n(n+1)}\right] = \tan^{-1}[x]$, then $x =$ (A) $\dfrac{1}{n+1}$ (B) $\dfrac{n}{n+1}$ (C) $\dfrac{1}{n+2}$ (D) $\dfrac{n}{n+2}$

   ›Reveal solutionSolution

   Each term telescopes as $\tan^{-1}(k+1)-\tan^{-1}(k)$; summing collapses the whole series to $\tan^{-1}\big(\frac{n}{n+2}\big)$, so $x=\dfrac{n}{n+2}$.

   Concept and Intuition

   The key identity is $\tan^{-1}\dfrac{1}{1+k(k+1)} = \tan^{-1}(k+1)-\tan^{-1}k$, which follows from the tangent subtraction formula $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\dfrac{p-q}{1+pq}$ applied with $p=k+1,\,q=k$. Once each term is in this "difference" form, the whole sum telescopes, leaving only the first and last pieces.

   Step-by-Step Solution

   1. Verify the telescoping identity: $\tan^{-1}(k+1)-\tan^{-1}(k) = \tan^{-1}\dfrac{(k+1)-k}{1+(k+1)k} = \tan^{-1}\dfrac{1}{1+k(k+1)}$. ✓ matches each term's form (with $k=1,2,\dots,n$).
   2. So the sum $\sum_{k=1}^{n}\tan^{-1}\dfrac{1}{1+k(k+1)} = \sum_{k=1}^{n}\left[\tan^{-1}(k+1)-\tan^{-1}(k)\right]$.
   3. This telescopes: all intermediate terms cancel, leaving $\tan^{-1}(n+1) - \tan^{-1}(1)$.
   4. Apply the subtraction formula again: $\tan^{-1}(n+1)-\tan^{-1}(1) = \tan^{-1}\dfrac{(n+1)-1}{1+(n+1)(1)} = \tan^{-1}\dfrac{n}{n+2}$. …

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9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Match the items of List - I with those of List - II List - I: A. $Tan^{-1}3 + Tan^{-1}x = Tan^{-1}8 \Rightarrow x =$ B. $Sin^{-1}x - Cos^{-1}x = \dfrac{\pi}{6} \Rightarrow x =$ C. $Sin^{-1}\dfrac{4}{5} + 2\,Tan^{-1}\dfrac{1}{3} =$ D. $\tan\left(Sec^{-1}\dfrac{1}{x}\right) = \sin(Tan^{-1}2), x > 0 \Rightarrow x =$ List - II: I. $\dfrac{\sqrt{5}}{3}$ II. $\dfrac{1}{5}$ III. $\dfrac{\sqrt{3}}{2}$ IV. $\dfrac{\pi}{2}$ V. $\dfrac{\pi}{3}$ The Correct Match is: (A) A-I, B-III, C-V, D-IV (B) A-II, B-III, C-IV, D-I (C) A-III, B-II, C-IV, D-V (D) A-II, B-I, C-IV, D-V

   ›Reveal solutionSolution

   This is a match-the-column on inverse trig identities. Answer: A-II, B-III, C-IV, D-I.

   Concept and Intuition

   Each item reduces via a standard inverse-trig identity: the tangent-addition formula, the complementary relation $\sin^{-1}x+\cos^{-1}x=\pi/2$, the double-angle formula for $\tan^{-1}$, and converting $\sec^{-1}$/$\tan^{-1}$ expressions into a right-triangle ratio.

   Step-by-Step Solution

   A. $\tan^{-1}3+\tan^{-1}x=\tan^{-1}8$. Take tangent of both sides: $\dfrac{3+x}{1-3x}=8 \Rightarrow 3+x=8-24x \Rightarrow 25x=5 \Rightarrow x=\dfrac15$. Matches II.

   B. $\sin^{-1}x-\cos^{-1}x=\dfrac{\pi}{6}$, and always $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$. Adding: $2\sin^{-1}x = \dfrac{\pi}{2}+\dfrac{\pi}{6}=\dfrac{2\pi}{3} \Rightarrow \sin^{-1}x=\dfrac{\pi}{3} \Rightarrow x=\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}$. Matches III.

   C. $\sin^{-1}\dfrac45 = \tan^{-1}\dfrac43$ (right triangle with opposite 4, hypotenuse 5, adjacent 3). Also $2\tan^{-1}\dfrac13$: using $\tan2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac34$, so $2\tan^{-1}\dfrac13=\tan^{-1}\dfrac34$. Sum $=\tan^{-1}\dfrac43+\tan^{-1}\dfrac34$; since $\dfrac43\times\dfrac34=1$ (reciprocal tangents), this sum is $\dfrac{\pi}{2}$. Matches IV. …

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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 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $\theta=\text{Tan}^{-1}\left(\dfrac{1}{3}\right)+\text{Tan}^{-1}\left(\dfrac{1}{7}\right)+\text{Tan}^{-1}\left(\dfrac{1}{13}\right)+\text{Tan}^{-1}\left(\dfrac{1}{21}\right)+\text{Tan}^{-1}\left(\dfrac{1}{31}\right)$, then $\tan\theta=$ (A) $\dfrac{3}{5}$ (B) 1 (C) $\dfrac{5}{7}$ (D) $\dfrac{7}{9}$

    ›Reveal solutionSolution

    Recognizing $3,7,13,21,31$ as the sequence $n^2+n+1$ turns each $\arctan$ term into a telescoping difference $\arctan(n+1)-\arctan n$. Answer: $\tan\theta=5/7$.

    Concept and Intuition

    The telescoping arctan identity $\arctan(n+1)-\arctan n=\arctan\dfrac{1}{1+n(n+1)}=\arctan\dfrac1{n^2+n+1}$ turns a long sum of small-angle arctangents into just the first and last terms — a huge simplification once the denominators are recognized as $n^2+n+1$.

    Step-by-Step Solution

    1. Check the denominators: for $n=1,2,3,4,5$, $n^2+n+1=3,7,13,21,31$ — exactly matching the given series.
    2. Use $\tan^{-1}(n+1)-\tan^{-1}n=\tan^{-1}\dfrac{(n+1)-n}{1+n(n+1)}=\tan^{-1}\dfrac{1}{n^2+n+1}$.
    3. So $\theta=\displaystyle\sum_{n=1}^{5}\left[\tan^{-1}(n+1)-\tan^{-1}n\right]$, which telescopes: all middle terms cancel, leaving $\theta=\tan^{-1}6-\tan^{-1}1$.
    4. $\tan^{-1}1=\pi/4$, so $\theta=\tan^{-1}6-\pi/4$. …

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12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.$\displaystyle\lim_{n \to \infty} \sum_{r=1}^{n} \cot^{-1}\left(r^2 + \dfrac{3}{4}\right) =$ (A) $\cot^{-1} 2$ (B) $\cot^{-1} \dfrac{1}{3}$ (C) $\tan^{-1} 2$ (D) $\tan^{-1} \dfrac{1}{3}$

    ›Reveal solutionSolution

    Rewriting each term as a difference of two arctangents makes the sum telescope; taking the limit as $n\to\infty$ gives $\tan^{-1}2$.

    Concept and Intuition

    Sums of $\tan^{-1}$ or $\cot^{-1}$ terms with a quadratic argument like $r^2+c$ often telescope, because $r^2+\tfrac34 = 1+(r+\tfrac12)(r-\tfrac12)$ matches the arctangent subtraction identity $\tan^{-1}A-\tan^{-1}B=\tan^{-1}\dfrac{A-B}{1+AB}$ with $A=r+\tfrac12$, $B=r-\tfrac12$ (so $A-B=1$).

    Step-by-Step Solution

    1. $\cot^{-1}\left(r^2+\tfrac34\right)=\tan^{-1}\dfrac{1}{r^2+3/4}$.
    2. Let $A=r+\tfrac12$, $B=r-\tfrac12$: then $AB=r^2-\tfrac14$, so $1+AB=r^2+\tfrac34$, and $A-B=1$.
    3. So $\tan^{-1}\dfrac{1}{r^2+3/4}=\tan^{-1}\dfrac{A-B}{1+AB}=\tan^{-1}A-\tan^{-1}B=\tan^{-1}\!\left(r+\tfrac12\right)-\tan^{-1}\!\left(r-\tfrac12\right)$.
    4. Sum from $r=1$ to $n$ telescopes: $\sum_{r=1}^{n}\left[\tan^{-1}(r+\tfrac12)-\tan^{-1}(r-\tfrac12)\right]=\tan^{-1}(n+\tfrac12)-\tan^{-1}(\tfrac12)$. …

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