Q.If $|x|\le1$, then $2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ is equal to
(A) $4\tan^{-1}x$
(B) $0$
(C) $\frac{\pi}{2}$
(D) $\pi$

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 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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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 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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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\sin\left(\text{Tan}^{-1}\dfrac{12}{17} + \text{Tan}^{-1}\dfrac{5}{29}\right) =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   This tests the sine-addition formula built from two inverse-tangent right triangles; the arithmetic collapses very neatly since $866=2\times433$.

   Concept and Intuition

   $\tan^{-1}\frac{12}{17}$ is the angle of a right triangle with opposite $12$, adjacent $17$, hypotenuse $\sqrt{12^2+17^2}=\sqrt{433}$. Similarly $\tan^{-1}\frac{5}{29}$ gives a triangle with hypotenuse $\sqrt{5^2+29^2}=\sqrt{866}$. Once we have $\sin$ and $\cos$ of both angles, the sum formula does the rest — no need to ever find the angles themselves.

   Step-by-Step Solution

   1. Let $A=\tan^{-1}\frac{12}{17}$: $\sin A=\dfrac{12}{\sqrt{433}}$, $\cos A=\dfrac{17}{\sqrt{433}}$ (since $12^2+17^2=144+289=433$).
   2. Let $B=\tan^{-1}\frac{5}{29}$: $\sin B=\dfrac{5}{\sqrt{866}}$, $\cos B=\dfrac{29}{\sqrt{866}}$ (since $5^2+29^2=25+841=866$).
   3. $\sin(A+B)=\sin A\cos B+\cos A\sin B=\dfrac{12(29)+17(5)}{\sqrt{433}\sqrt{866}}=\dfrac{348+85}{\sqrt{433\cdot866}}=\dfrac{433}{\sqrt{433\cdot866}}$. …

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6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\sin\left[2\tan^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{3}{5}\right)\right] =$ (A) 0 (B) 1 (C) $\dfrac{1}{2}$ (D) $\dfrac{\sqrt3}{2}$

   ›Reveal solutionSolution

   Convert both inverse-trig terms into a single angle's sine/cosine using right-triangle ratios, then apply the sine addition formula; the sum evaluates neatly to $1$.

   Concept and Intuition

   When adding two inverse trig angles, the cleanest approach is to name each one, extract its sine and cosine from the implied right triangle, then use the standard addition formula $\sin(A+B)=\sin A\cos B+\cos A\sin B$ — never try to add the angles numerically.

   Step-by-Step Solution

   1. Let $\varphi = \tan^{-1}\left(\dfrac12\right)$. In a right triangle, opposite $=1$, adjacent $=2$, hypotenuse $=\sqrt5$.
   2. Use the double-angle identity $\tan2\varphi = \dfrac{2\tan\varphi}{1-\tan^2\varphi} = \dfrac{2(1/2)}{1-(1/4)} = \dfrac{1}{3/4}=\dfrac43$.
   3. Since $\varphi\in(0,\pi/4)$ (as $\tan\varphi=1/2<1$), $2\varphi\in(0,\pi/2)$, so this is a genuine first-quadrant angle: with opposite $4$, adjacent $3$, hypotenuse $5$, giving $\sin2\varphi=4/5,\ \cos2\varphi=3/5$.
   4. Let $\psi=\sin^{-1}(3/5)$, so $\sin\psi=3/5$ and (principal branch, first quadrant) $\cos\psi=4/5$. …

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7. 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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8. 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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9. 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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10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $y = \mathrm{Tan}^{-1}\left(\dfrac{x}{1+2x^2}\right) + \mathrm{Tan}^{-1}\left(\dfrac{x}{1+6x^2}\right)$, then $\dfrac{dy}{dx}=$ (A) $\dfrac{4}{16x^2+1} - \dfrac{3}{9x^2+1}$ (B) $\dfrac{3}{9x^2+1} - \dfrac{1}{x^2+1}$ (C) $\dfrac{3}{9x^2+1} - \dfrac{2}{4x^2+1}$ (D) $\dfrac{1}{9x^2+1} - \dfrac{1}{x^2+1}$

    ›Reveal solutionSolution

    Recognizing each arctan term as a telescoping difference $\tan^{-1}(2x)-\tan^{-1}(x)$ and $\tan^{-1}(3x)-\tan^{-1}(2x)$ collapses $y$ to $\tan^{-1}3x-\tan^{-1}x$, whose derivative is immediate.

    Concept and Intuition

    Terms of the form $\tan^{-1}\!\left(\dfrac{a-b}{1+ab}\right)$ are exactly $\tan^{-1}a - \tan^{-1}b$ (the tangent subtraction identity, valid when $ab>-1$). Spotting this pattern turns an awkward-looking sum into a telescoping simplification, avoiding messy direct differentiation of nested rational-argument arctans.

    Step-by-Step Solution

    1. Compare $\dfrac{x}{1+2x^2}$ to the form $\dfrac{a-b}{1+ab}$: try $a=2x,\,b=x$, giving $\dfrac{2x-x}{1+2x\cdot x}=\dfrac{x}{1+2x^2}$ ✓. So

    $\tan^{-1}\!\left(\frac{x}{1+2x^2}\right) = \tan^{-1}(2x)-\tan^{-1}(x)$

    2. Compare $\dfrac{x}{1+6x^2}$ similarly: try $a=3x,\,b=2x$, giving $\dfrac{3x-2x}{1+3x\cdot2x}=\dfrac{x}{1+6x^2}$ ✓. So

    $\tan^{-1}\!\left(\frac{x}{1+6x^2}\right) = \tan^{-1}(3x)-\tan^{-1}(2x)$

    3. Add the two:

    $y = \big[\tan^{-1}2x-\tan^{-1}x\big] + \big[\tan^{-1}3x-\tan^{-1}2x\big] = \tan^{-1}3x - \tan^{-1}x$

    (the $\tan^{-1}2x$ terms cancel — a telescoping sum).

    4. Differentiate: …

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11. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.If the equation $2\cot^{-1}(x^2+2x+k) = \pi - 3\tan^{-1}(x^2+2x+k)$ has two distinct real solutions, then all the values of $k$ lie in the interval (A) $(-1, 2)$ (B) $(1, \infty)$ (C) $(-\infty, \infty)$ (D) $(-\infty, 1)$

    ›Reveal solutionSolution

    The inverse-trig equation collapses, via the identity $\cot^{-1}t+\tan^{-1}t=\pi/2$, into the purely algebraic condition $x^2+2x+k=0$; the question then just asks when this quadratic has two distinct real roots.

    Concept and Intuition

    $\tan^{-1}t$ and $\cot^{-1}t$ are complementary for every real $t$: $\cot^{-1}t=\pi/2-\tan^{-1}t$. Substituting this converts a mixed inverse-trig equation into a single equation in $\tan^{-1}t$ alone, which resolves to a specific numeric value of $t$. Once $t$ is pinned to a number, the "two distinct real solutions" condition is just the familiar discriminant test on the quadratic $t(x)=x^2+2x+k$.

    Step-by-Step Solution

    1. Let $t=x^2+2x+k$ (real for every real $x$).
    2. Given: $2\cot^{-1}t=\pi-3\tan^{-1}t$.
    3. Substitute $\cot^{-1}t=\dfrac{\pi}{2}-\tan^{-1}t$: $2\left(\dfrac{\pi}{2}-\tan^{-1}t\right)=\pi-3\tan^{-1}t$.
    4. Expand: $\pi-2\tan^{-1}t=\pi-3\tan^{-1}t$.
    5. Cancel $\pi$: $-2\tan^{-1}t=-3\tan^{-1}t\Rightarrow \tan^{-1}t=0\Rightarrow t=0$. …

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12. 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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