Q.If $y=2\tan^{-1}x+\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for all $x$, then ____ $<y<$ ____.

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 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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3. 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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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 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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6. 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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7. 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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8. 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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9. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $y = \tan^{-1}\left(\dfrac{a\cos x - b\sin x}{b\cos x + a\sin x}\right)$, then $\dfrac{dy}{dx} =$ ______ (A) $0$ (B) $\dfrac{a}{b}$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   Tests recognizing an $a\cos x - b\sin x$ / $b\cos x + a\sin x$ ratio as a shifted cotangent using the auxiliary-angle (R-method) substitution, collapsing the arctan to a linear function of $x$.

   Concept and Intuition

   Expressions like $a\cos x - b\sin x$ can always be written as $r\cos(x+\phi)$ for suitable $r=\sqrt{a^2+b^2}$ and angle $\phi$ with $a=r\cos\phi$, $b=r\sin\phi$. Once both numerator and denominator are expressed this way, their ratio collapses to a simple trig ratio in $(x+\phi)$, and the $\arctan$ of that becomes an explicit linear expression in $x$ — making differentiation trivial.

   Step-by-Step Solution

   1. Let $a=r\cos\phi$, $b=r\sin\phi$ where $r=\sqrt{a^2+b^2}$.
   2. Numerator: $a\cos x - b\sin x = r\cos\phi\cos x - r\sin\phi\sin x = r\cos(x+\phi)$.
   3. Denominator: $b\cos x + a\sin x = r\sin\phi\cos x + r\cos\phi\sin x = r\sin(x+\phi)$.
   4. Ratio: $\dfrac{r\cos(x+\phi)}{r\sin(x+\phi)}=\cot(x+\phi)=\tan\left(\dfrac\pi2-(x+\phi)\right)$. …

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10. 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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11. 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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12. 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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