Q.Find the values of each of the following: $\tan^{-1} \left(\frac{3a^2x-x^3}{a^3-3ax^2}\right)$, $a > 0$; $-\frac{a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$

Showing the 12 most recent of 34 on this concept.

1. 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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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 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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4. 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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5. 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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6. 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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7. 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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8. 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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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 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If $y = \mathrm{Tan}^{-1}\left\{\dfrac{ax-b}{bx+a}\right\}$, then $y' =$ _______ (A) $\dfrac{1}{1+x^2} + \dfrac{a^2}{a^2+b^2}$ (B) $\dfrac{1}{1+x^2}$ (C) $\dfrac{1}{1+\left(\dfrac{ax-b}{bx+a}\right)^2}$ (D) $\dfrac{bx+a}{1+(ax-b)^2}$

    ›Reveal solutionSolution

    This tests differentiating an arctangent of a Möbius-type expression, which (as often happens) collapses to the simple form $\tfrac{1}{1+x^2}$. Answer: $\dfrac{1}{1+x^2}$.

    Concept and Intuition

    Expressions of the form $\mathrm{Tan}^{-1}\!\left(\dfrac{ax-b}{bx+a}\right)$ often equal $\mathrm{Tan}^{-1}(x) - \mathrm{Tan}^{-1}(b/a)$ up to a constant (since $\dfrac{ax-b}{bx+a} = \dfrac{\frac{a}{b}x - 1}{x+\frac ab}$ resembles $\tan(\alpha-\beta)$ with a constant angle $\beta$), so its derivative should reduce to just $\dfrac{1}{1+x^2}$ — the constant angle contributes zero derivative.

    Step-by-Step Solution

    1. Let $u = \dfrac{ax-b}{bx+a}$. By the quotient rule, $u' = \dfrac{a(bx+a) - (ax-b)b}{(bx+a)^2} = \dfrac{abx+a^2-abx+b^2}{(bx+a)^2} = \dfrac{a^2+b^2}{(bx+a)^2}$.
    2. Compute $1+u^2 = 1+\dfrac{(ax-b)^2}{(bx+a)^2} = \dfrac{(bx+a)^2+(ax-b)^2}{(bx+a)^2}$. …

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11. 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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12. 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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