Q.Find the value of the following: $\cot^{-1} \left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}$, $x \in \left(0, \frac{\pi}{4}\right)$

Showing the 12 most recent of 34 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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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 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $y = \log_{\cot x} \tan x - \log_{\tan x} \cot x + \tan^{-1}\left(\dfrac{4x}{4-x^2}\right)$, then $\dfrac{dy}{dx} =$ ______ (A) $\dfrac{1}{4+x^2}$ (B) $\dfrac{4}{4+x^2}$ (C) $\dfrac{1}{4-x^2}$ (D) $\dfrac{4}{4-x^2}$

   ›Reveal solutionSolution

   This tests the change-of-base log identity and the double-angle form of $\tan^{-1}$. The two log terms cancel completely, and the answer is $\dfrac{4}{4+x^2}$.

   Concept and Intuition

   Whenever you see $\log_a b$ and $\log_b a$ together, remember they are reciprocals of each other: $\log_a b=1/\log_b a$. Here $a=\cot x,\,b=\tan x$ are reciprocals of each other too, so $\ln(\tan x)=-\ln(\cot x)$, which forces both log terms to equal $-1$ and cancel. What's left is the classic $\tan^{-1}\left(\dfrac{2t}{1-t^2}\right)=2\tan^{-1}t$ substitution pattern with $t=x/2$.

   Step-by-Step Solution

   1. $\log_{\cot x}\tan x=\dfrac{\ln\tan x}{\ln\cot x}=\dfrac{\ln\tan x}{\ln(1/\tan x)}=\dfrac{\ln\tan x}{-\ln\tan x}=-1$.
   2. Similarly $\log_{\tan x}\cot x=-1$.
   3. So $y=-1-(-1)+\tan^{-1}\!\left(\dfrac{4x}{4-x^2}\right)=\tan^{-1}\!\left(\dfrac{4x}{4-x^2}\right)$. …

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9. 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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10. 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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11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.For how many distinct values of $x$, the following $\sin[2\cos^{-1}\cot(2\tan^{-1}x)] = 0$ holds? (A) 8 (B) 2 (C) 6 (D) 4

    ›Reveal solutionSolution

    Substituting $\theta=\tan^{-1}x$ reduces the equation to finding where $\cot2\theta\in\{-1,0,1\}$, giving exactly 6 distinct values of $x$.

    Concept and Intuition

    When an equation is built from a chain of inverse trig functions, it's cleanest to work from the outside in: first find which values of the innermost composite expression make the whole thing zero, then solve backward.

    Step-by-Step Solution

    1. Let $\theta=\tan^{-1}x\in(-\pi/2,\pi/2)$, so $2\theta\in(-\pi,\pi)$, and let $u=\cot(2\theta)$ (needs $\theta\ne0$).
    2. Need $\cos^{-1}(u)\in[0,\pi]$ to satisfy $\sin[2\cos^{-1}(u)]=0$, i.e. $2\cos^{-1}(u)=k\pi$ for $k=0,1,2$ (since $\cos^{-1}(u)\in[0,\pi]$ bounds $k$ to these three).
    3. $k=0\Rightarrow\cos^{-1}(u)=0\Rightarrow u=1$. $k=1\Rightarrow\cos^{-1}(u)=\pi/2\Rightarrow u=0$. $k=2\Rightarrow\cos^{-1}(u)=\pi\Rightarrow u=-1$.
    4. So need $\cot(2\theta)\in\{-1,0,1\}$ with $2\theta\in(-\pi,\pi)\setminus\{0\}$.
    5. $\cot(2\theta)=0$: $2\theta=\pm\pi/2\Rightarrow\theta=\pm\pi/4$ — 2 solutions. …

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12. 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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