Q.If $\sin^{-1}\left(\frac{2a}{1+a^2}\right)+\cos^{-1}\left(\frac{1-a^2}{1+a^2}\right)=\tan^{-1}\left(\frac{2x}{1-x^2}\right)$, where $a, x\in\,]0,1[$, then the value of $x$ is
(A) $0$
(B) $\frac{a}{2}$
(C) $a$
(D) $\frac{2a}{1-a^2}$

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. $\phi=\tan^{-1}\!\left(\dfrac{\tan\theta}{3}\right)$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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: $\dfrac{x^2}{1-x^2} = \dfrac{b^2}{b^2-a^2}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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: $\tan^{-1}(n+1)-\tan^{-1}(1) = \tan^{-1}\dfrac{(n+1)-1}{1+(n+1)(1)} = \tan^{-1}\dfrac{n}{n+2}$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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). …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $y=\tan^{-1}\!\left[\tan\!\left(\dfrac\pi4+\dfrac\varphi2\right)\right]=\dfrac\pi4+\dfrac\varphi2=\dfrac\pi4+\dfrac12\cos^{-1}(x^2)$. …

   🔒 Start your 14-day free trial to see the full solution →
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: $\dfrac{1-x^2}{1+x^2}=\dfrac{24}{25}$. …

   🔒 Start your 14-day free trial to see the full solution →
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: $x\,\mathrm{Cot}^{-1}x+\log\sqrt{1+x^2}+c$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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) $\cos\psi=4/5$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If $y=\text{Tan}^{-1}\dfrac{x}{1+2x^2}+\text{Tan}^{-1}\dfrac{x}{1+6x^2}+\text{Tan}^{-1}\dfrac{x}{1+12x^2}$, then $\left(\dfrac{dy}{dx}\right)_{x=\frac{1}{2}}=$ (A) $1$ (B) $-1$ (C) $0$ (D) $\frac{1}{2}$

    ›Reveal solutionSolution

    Recognize the telescoping arctan pattern to collapse $y$ to $\arctan(4x)-\arctan(x)$; the derivative at $x=\frac12$ is $0$.

    Concept and Intuition

    The identity $\arctan A-\arctan B=\arctan\dfrac{A-B}{1+AB}$ (mod branch issues) means a sum like $\arctan\dfrac{x}{1+n(n+1)x^2}$, recognized as $\arctan((n+1)x)-\arctan(nx)$, telescopes when summed over consecutive $n$ — a huge simplification before ever differentiating.

    Step-by-Step Solution

    1. Check the general term: $\arctan((n+1)x)-\arctan(nx)=\arctan\dfrac{(n+1)x-nx}{1+n(n+1)x^2}=\arctan\dfrac{x}{1+n(n+1)x^2}$.
    2. Match given terms: $\dfrac{x}{1+2x^2}$ has $n(n+1)=2\Rightarrow n=1$; $\dfrac{x}{1+6x^2}$ has $n(n+1)=6\Rightarrow n=2$; $\dfrac{x}{1+12x^2}$ has $n(n+1)=12\Rightarrow n=3$.
    3. So $y=[\arctan2x-\arctan x]+[\arctan3x-\arctan2x]+[\arctan4x-\arctan3x]=\arctan4x-\arctan x$ (everything else cancels). …

    🔒 Start your 14-day free trial to see the full solution →
12. 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. $\sin(A+B)=\sin A\cos B+\cos A\sin B=\dfrac{12(29)+17(5)}{\sqrt{433}\sqrt{866}}=\dfrac{348+85}{\sqrt{433\cdot866}}=\dfrac{433}{\sqrt{433\cdot866}}$. …

    🔒 Start your 14-day free trial to see the full solution →