Q.Prove that $\tan^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^2$.

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-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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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 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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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-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). …

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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 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 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$4\tan^{-1}\dfrac{1}{5} - \tan^{-1}\dfrac{1}{70} + \tan^{-1}\dfrac{1}{99} =$ (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   This is a Machin-like arctangent identity that evaluates exactly to $\pi/4$.

   Concept and Intuition

   Sums/differences of $\tan^{-1}$ terms with small reciprocal arguments often combine (via repeated use of $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\frac{p-q}{1+pq}$ and the double/quadruple-angle formula for tangent) into a single nice angle like $\pi/4$. These are the classical 'Machin-type' formulas historically used to compute $\pi$.

   Step-by-Step Solution

   1. First combine $4\tan^{-1}\frac15$ using the double-angle formula for $\tan$ twice: with $\tan\alpha=\frac15$, $\tan2\alpha=\dfrac{2\cdot\frac15}{1-\frac1{25}}=\dfrac{2/5}{24/25}=\dfrac{5}{12}$, and $\tan4\alpha=\dfrac{2\cdot\frac5{12}}{1-\frac{25}{144}}=\dfrac{5/6}{119/144}=\dfrac{120}{119}$.
   2. So $4\tan^{-1}\frac15 = \tan^{-1}\dfrac{120}{119}$ (in the correct quadrant, since $\frac{120}{119}$ is only slightly bigger than 1, the angle is just over $\pi/4$).
   3. Now combine $\tan^{-1}\dfrac{120}{119}-\tan^{-1}\dfrac1{70}$ using $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\frac{p-q}{1+pq}$: numerator $\frac{120}{119}-\frac1{70} = \frac{120\cdot70-119}{119\cdot70}=\frac{8400-119}{8330}=\frac{8281}{8330}$; denominator $1+\frac{120}{119\cdot70}=1+\frac{120}{8330}=\frac{8450}{8330}$. Ratio $=\frac{8281}{8450}$. …

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10. 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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11. 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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12. 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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