Q.Prove that: $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1}\left(\frac{1-x}{1+x}\right)$, $x \in [0, 1]$.

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $y = \tan^{-1}\left(\dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}}\right)$, where $x^2 \le 1$. Then find $\dfrac{dy}{dx}$ (A) $\dfrac{\pi}{4} + \dfrac{1}{2}\cos^{-1}(x^2)$ (B) $\dfrac{\pi}{4} - \dfrac{1}{2}\cos^{-1}(x^2)$ (C) $\dfrac{-x}{\sqrt{1-x^4}}$ (D) $\dfrac{-2x}{\sqrt{1-x^4}}$

   ›Reveal solutionSolution

   Substituting $x^2=\cos\varphi$ collapses the expression to $y=\pi/4+\tfrac12\cos^{-1}(x^2)$, whose derivative is $\dfrac{-x}{\sqrt{1-x^4}}$.

   Concept and Intuition

   Nested-radical inverse-trig expressions like this are almost always designed to simplify via a trig substitution that turns $\sqrt{1\pm x^2}$ into $\sqrt2\cos$ or $\sqrt2\sin$ of a half-angle, converting the whole ratio into a single tangent — much easier to differentiate than the raw radical expression.

   Step-by-Step Solution

   1. Let $x^2=\cos\varphi$ (valid since $x^2\le1$). Then $1+x^2=2\cos^2(\varphi/2)$ and $1-x^2=2\sin^2(\varphi/2)$.
   2. So $\sqrt{1+x^2}=\sqrt2\cos(\varphi/2)$ and $\sqrt{1-x^2}=\sqrt2\sin(\varphi/2)$.
   3. The ratio becomes $\dfrac{\cos(\varphi/2)+\sin(\varphi/2)}{\cos(\varphi/2)-\sin(\varphi/2)} = \dfrac{1+\tan(\varphi/2)}{1-\tan(\varphi/2)} = \tan\!\left(\dfrac{\pi}{4}+\dfrac{\varphi}{2}\right)$.
   4. So $y=\tan^{-1}\!\left[\tan\!\left(\dfrac\pi4+\dfrac\varphi2\right)\right]=\dfrac\pi4+\dfrac\varphi2=\dfrac\pi4+\dfrac12\cos^{-1}(x^2)$. …

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2. 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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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The number of real solutions of the equation $\tan^{-1}\sqrt{x(x+1)} + \sin^{-1}\sqrt{x^2+x+1} = \dfrac{\pi}{2}$ is (A) $0$ (B) $1$ (C) $2$ (D) Infinitely many

   ›Reveal solutionSolution

   The domain restrictions of $\tan^{-1}\sqrt{\cdot}$ and $\sin^{-1}\sqrt{\cdot}$ force $x(x+1)=0$, giving exactly two real solutions, $x=0$ and $x=-1$.

   Concept and Intuition

   Before manipulating an inverse-trig equation, always pin down the domain first — here the two square roots and the $\sin^{-1}$ range constraint do almost all the work.

   Step-by-Step Solution

   1. Let $y=x^2+x=x(x+1)$. For $\tan^{-1}\sqrt y$ to be real we need $y\ge0$.
   2. Note $x^2+x+1=y+1$. For $\sin^{-1}\sqrt{y+1}$ to be defined we need $0\le y+1\le1$, i.e. $-1\le y\le0$.
   3. Combining $y\ge0$ and $y\le0$ forces $y=0$ exactly. …

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4. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $y=\text{Tan}^{-1}\sqrt{\dfrac{1-\cos x}{1+\cos x}}$, then the values of $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$ respectively are ________ (A) $1,\ 0$ (B) $\dfrac{x}{2},\ \dfrac{1}{2}$ (C) $\dfrac{1}{2},\ 0$ (D) $\dfrac{-1}{2},\ 0$

   ›Reveal solutionSolution

   The half-angle identity collapses $y$ to $x/2$, so $dy/dx = 1/2$ and $d^2y/dx^2 = 0$.

   Concept and Intuition

   Expressions like $\sqrt{\dfrac{1-\cos x}{1+\cos x}}$ are classic half-angle simplifications; recognizing the identity turns an intimidating inverse-trig derivative problem into a trivial linear function.

   Step-by-Step Solution

   1. Recall $1-\cos x = 2\sin^2(x/2)$ and $1+\cos x = 2\cos^2(x/2)$.
   2. So $\dfrac{1-\cos x}{1+\cos x} = \tan^2(x/2)$, and $\sqrt{\tan^2(x/2)} = |\tan(x/2)|$.
   3. On the principal branch where $\tan(x/2)\ge 0$ (i.e. $x\in(-\pi,\pi)$), $y = \operatorname{Tan}^{-1}(\tan(x/2)) = \dfrac{x}{2}$.
   4. Differentiate: $\dfrac{dy}{dx} = \dfrac12$.
   5. Differentiate again: since $dy/dx$ is constant, $\dfrac{d^2y}{dx^2} = 0$.

   Common Mistakes …

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5. 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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6. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.The number of solutions of $\operatorname{Tan}^{-1}1+\dfrac{1}{2}\operatorname{Cos}^{-1}x^2-\operatorname{Tan}^{-1}\left(\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=0$ is (A) 3 (B) 0 (C) 1 (D) infinitely many

   ›Reveal solutionSolution

   This tests simplifying an inverse-trig expression via a trigonometric substitution, revealing that the equation is actually an identity over its whole domain. Answer: infinitely many solutions.

   Concept and Intuition

   The fraction $\dfrac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ looks intimidating, but substituting $x^2=\cos\alpha$ converts $1\pm x^2$ into $2\cos^2(\alpha/2)$ and $2\sin^2(\alpha/2)$, turning the whole fraction into a clean $\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)$. This is the standard trick for expressions of the form $\sqrt{1+x^2}\pm\sqrt{1-x^2}$.

   Step-by-Step Solution

   1. Domain: $\operatorname{Cos}^{-1}(x^2)$ needs $x^2\in[-1,1]$, and since $x^2\ge0$ always, effectively $x^2\in[0,1]$, i.e. $x\in[-1,1]$. Also need $\sqrt{1-x^2}$ real, consistent.
   2. Let $\alpha=\operatorname{Cos}^{-1}(x^2)\in[0,\pi/2]$ (since $x^2\in[0,1]$, $\alpha$ can only range over $[0,\pi/2]$, not the full $[0,\pi]$).
   3. Then $x^2=\cos\alpha$, so $1+x^2=1+\cos\alpha=2\cos^2(\alpha/2)$ and $1-x^2=1-\cos\alpha=2\sin^2(\alpha/2)$. Since $\alpha/2\in[0,\pi/4]$, both $\cos(\alpha/2),\sin(\alpha/2)\ge0$, so $\sqrt{1+x^2}=\sqrt2\cos(\alpha/2)$, $\sqrt{1-x^2}=\sqrt2\sin(\alpha/2)$.
   4. The fraction becomes $\dfrac{\cos(\alpha/2)+\sin(\alpha/2)}{\cos(\alpha/2)-\sin(\alpha/2)} = \dfrac{1+\tan(\alpha/2)}{1-\tan(\alpha/2)} = \tan\left(\dfrac{\pi}{4}+\dfrac{\alpha}{2}\right)$.
   5. Since $\alpha/2\in[0,\pi/4]$, we have $\dfrac{\pi}{4}+\dfrac{\alpha}{2}\in\left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$, safely inside the principal range of $\operatorname{Tan}^{-1}$, so $\operatorname{Tan}^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right]=\dfrac{\pi}{4}+\dfrac{\alpha}{2}$ exactly (excluding $x=0$ where $\alpha=\pi/2$ makes the denominator zero). …

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7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.For $0 < x < 1$, $\displaystyle\int [\mathrm{Tan}^{-1}(1 - x + x^2) + \mathrm{Tan}^{-1}(1 - x)]\, dx =$ (A) $x\,\mathrm{Cot}^{-1}x + \log\sqrt{1+x^2} + c$ (B) $x\,\mathrm{Tan}^{-1}x - \log(1+x^2) + c$ (C) $x\,\mathrm{Cot}^{-1}x + \dfrac{3}{4}\log(1+x^2) + c$ (D) $x\,\mathrm{Tan}^{-1}x - \dfrac{3}{4}\log\sqrt{1+x^2} + c$

   ›Reveal solutionSolution

   The two arctangent terms combine, via the tan-addition identity, into a single $\mathrm{Cot}^{-1}x$; integrating that by parts gives option (A).

   Concept and Intuition

   The stem looks intimidating because it has two separate inverse-tangent terms with messy arguments. The key insight is that $\mathrm{Tan}^{-1}p+\mathrm{Tan}^{-1}q$ always collapses via

   $\mathrm{Tan}^{-1}p+\mathrm{Tan}^{-1}q=\mathrm{Tan}^{-1}\!\left(\frac{p+q}{1-pq}\right)\ (\text{mod }\pi\text{ correction}),$

   so it's worth testing whether $p=1-x+x^2$ and $q=1-x$ are designed to make $\dfrac{p+q}{1-pq}$ simplify beautifully — which they are.

   Step-by-Step Solution

   1. Compute $p+q = (1-x+x^2)+(1-x) = 2-2x+x^2$.
   2. Compute $pq = (1-x+x^2)(1-x)$. Expanding: $(1-x+x^2)(1-x) = 1-2x+2x^2-x^3$.
   3. So $1-pq = 1-(1-2x+2x^2-x^3) = 2x-2x^2+x^3 = x(2-2x+x^2)$.
   4. Hence $\dfrac{p+q}{1-pq} = \dfrac{2-2x+x^2}{x(2-2x+x^2)} = \dfrac{1}{x}$.
   5. Check the correction term: for $0<x<1$, $2-2x+x^2=(x-1)^2+1>0$ and $x>0$, so $1-pq>0 \Rightarrow pq<1$, meaning the plain addition formula applies with no $\pm\pi$ shift.
   6. So the integrand is exactly $\mathrm{Tan}^{-1}(1/x) = \mathrm{Cot}^{-1}x$ (valid since $x>0$).
   7. Now integrate by parts: $\displaystyle\int \mathrm{Cot}^{-1}x\,dx = x\,\mathrm{Cot}^{-1}x - \int x\cdot\left(\frac{-1}{1+x^2}\right)dx = x\,\mathrm{Cot}^{-1}x+\int\frac{x}{1+x^2}dx$.
   8. $\displaystyle\int\frac{x}{1+x^2}dx = \frac12\log(1+x^2)=\log\sqrt{1+x^2}$.
   9. Total: $x\,\mathrm{Cot}^{-1}x+\log\sqrt{1+x^2}+c$. …

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8. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The value of $x$ such that $\sin\left(2\tan^{-1}\dfrac{3}{4}\right) = \cos(2\tan^{-1}x)$ is (A) $7$ (B) $\dfrac{3}{7}$ (C) $\dfrac{1}{7}$ (D) $\dfrac{4}{7}$

   ›Reveal solutionSolution

   This tests the tangent double-angle formulas for both sine and cosine. Answer: $x=1/7$.

   Concept and Intuition

   Both sides are "double angle of an inverse tangent," so express each side purely in terms of the tangent using $\sin2\theta=\dfrac{2t}{1+t^2}$ and $\cos2\phi=\dfrac{1-x^2}{1+x^2}$ (both derivable from a right triangle with opposite/adjacent $=t$ or $x$).

   Step-by-Step Solution

   1. Let $\theta=\tan^{-1}(3/4)$, so $\tan\theta=3/4$.
   2. $\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}=\dfrac{2(3/4)}{1+9/16}=\dfrac{3/2}{25/16}=\dfrac{3}{2}\cdot\dfrac{16}{25}=\dfrac{24}{25}$.
   3. Let $\phi=\tan^{-1}x$, so $\cos2\phi=\dfrac{1-x^2}{1+x^2}$.
   4. Equation: $\dfrac{1-x^2}{1+x^2}=\dfrac{24}{25}$. …

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9. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.For $b > a$, the solution of the equation $\cot(\cos^{-1}x) = \sec\left\{\tan^{-1}\dfrac{a}{\sqrt{b^2-a^2}}\right\}$ is (A) $\dfrac{b}{\sqrt{2b^2-a^2}}$ (B) $\dfrac{a}{\sqrt{2b^2-a^2}}$ (C) $\dfrac{\sqrt{b^2-a^2}}{a}$ (D) $\dfrac{\sqrt{b^2-a^2}}{2b}$

   ›Reveal solutionSolution

   This tests converting inverse trig expressions into right-triangle ratios and solving the resulting algebraic equation. The answer is $x=\dfrac{b}{\sqrt{2b^2-a^2}}$.

   Concept and Intuition

   For any inverse trig expression like $\tan^{-1}(p/q)$, imagine a right triangle with opposite $p$ and adjacent $q$; the hypotenuse follows from the Pythagorean theorem, and any other trig ratio of that same angle can then be read straight off the triangle. Applying this to both sides converts the equation into pure algebra in $x$.

   Step-by-Step Solution

   1. Let $\theta = \tan^{-1}\dfrac{a}{\sqrt{b^2-a^2}}$. In the corresponding right triangle: opposite $=a$, adjacent $=\sqrt{b^2-a^2}$, so hypotenuse $=\sqrt{a^2+(b^2-a^2)}=b$.
   2. Hence $\sec\theta = \dfrac{\text{hyp}}{\text{adj}} = \dfrac{b}{\sqrt{b^2-a^2}}$.
   3. Let $\varphi=\cos^{-1}x$, so $\cos\varphi=x$: adjacent $=x$, hypotenuse $=1$, opposite $=\sqrt{1-x^2}$.
   4. Hence $\cot\varphi = \dfrac{\cos\varphi}{\sin\varphi} = \dfrac{x}{\sqrt{1-x^2}}$.
   5. Equating: $\dfrac{x}{\sqrt{1-x^2}} = \dfrac{b}{\sqrt{b^2-a^2}}$. Squaring: $\dfrac{x^2}{1-x^2} = \dfrac{b^2}{b^2-a^2}$. …

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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-2025-05-21-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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12. 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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