Q.$\tan^{-1} \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$ [Hint: Put $x = \cos 2\theta$] Solve the following equations:

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-2021-08-19-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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3. 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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4. 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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5. 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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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 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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8. 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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9. 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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10. 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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11. 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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12. 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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