Q.If $2\tan^{-1}(\cos\theta)=\tan^{-1}(2\csc\theta)$, then show that $\theta=\frac{\pi}{4}$.

Showing the 12 most recent of 34 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.$\text{Tan}^{-1}(-2) - \text{Tan}^{-1}(3)$ is equal to (A) $\dfrac{3\pi}{4}$ (B) $\dfrac{-\pi}{6}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{-3\pi}{4}$

   ›Reveal solutionSolution

   Convert to $-(\tan^{-1}2+\tan^{-1}3)$ and apply the addition formula (with the $+\pi$ correction since $ab>1$) to get $-\frac{3\pi}{4}$.

   Concept and Intuition

   $\tan^{-1}$ is an odd function, so $\tan^{-1}(-x)=-\tan^{-1}(x)$. The standard addition formula $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\frac{a+b}{1-ab}$ needs a $+\pi$ correction whenever $a,b>0$ and $ab>1$, because then the true sum exceeds $\pi/2$ while the raw arctan formula would return a negative principal value.

   Step-by-Step Solution

   1. $\tan^{-1}(-2) = -\tan^{-1}(2)$, so the expression becomes $-\tan^{-1}(2)-\tan^{-1}(3)$.
   2. Compute $\tan^{-1}2+\tan^{-1}3$. Here $a=2,b=3$, $ab=6>1$, both positive, so use $\tan^{-1}a+\tan^{-1}b = \pi+\tan^{-1}\frac{a+b}{1-ab}$.
   3. $\frac{a+b}{1-ab} = \frac{5}{1-6}=\frac{5}{-5}=-1$, and $\tan^{-1}(-1)=-\frac{\pi}{4}$. …

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7. 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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8. 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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9. 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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10. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.What is the value of $\mathrm{Sin}^{-1}\dfrac{12}{13} + \mathrm{Cos}^{-1}\dfrac{4}{5} + \mathrm{Tan}^{-1}\dfrac{63}{16} =$ (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{3\pi}{4}$

    ›Reveal solutionSolution

    The three inverse-trig terms telescope to exactly $\pi$ because the first two sum to $180^\circ$ minus the third.

    Concept and Intuition

    $\sin^{-1}\frac{12}{13}$ is the acute angle with sine $\frac{12}{13}$, cosine $\frac{5}{13}$, tangent $\frac{12}{5}$. $\cos^{-1}\frac{4}{5}$ is the acute angle with cosine $\frac{4}{5}$, sine $\frac{3}{5}$, tangent $\frac{3}{4}$. Adding two acute angles whose tangent-sum formula gives a negative tangent tells us their sum exceeds $90^\circ$ — a classic trick for handling sums of inverse trig terms without a calculator.

    Step-by-Step Solution

    1. Let $\alpha=\sin^{-1}\frac{12}{13}$ (so $\tan\alpha=\frac{12}{5}$) and $\beta=\cos^{-1}\frac{4}{5}$ (so $\tan\beta=\frac{3}{4}$).
    2. $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}=\dfrac{\frac{63}{20}}{-\frac{16}{20}}=-\dfrac{63}{16}$.
    3. Since $\alpha\approx 67.4^\circ,\beta\approx 36.9^\circ$, their sum $\alpha+\beta\approx 104.3^\circ\in(90^\circ,180^\circ)$, where tangent is negative — consistent. …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $\theta=\text{Tan}^{-1}\left(\dfrac{1}{3}\right)+\text{Tan}^{-1}\left(\dfrac{1}{7}\right)+\text{Tan}^{-1}\left(\dfrac{1}{13}\right)+\text{Tan}^{-1}\left(\dfrac{1}{21}\right)+\text{Tan}^{-1}\left(\dfrac{1}{31}\right)$, then $\tan\theta=$ (A) $\dfrac{3}{5}$ (B) 1 (C) $\dfrac{5}{7}$ (D) $\dfrac{7}{9}$

    ›Reveal solutionSolution

    Recognizing $3,7,13,21,31$ as the sequence $n^2+n+1$ turns each $\arctan$ term into a telescoping difference $\arctan(n+1)-\arctan n$. Answer: $\tan\theta=5/7$.

    Concept and Intuition

    The telescoping arctan identity $\arctan(n+1)-\arctan n=\arctan\dfrac{1}{1+n(n+1)}=\arctan\dfrac1{n^2+n+1}$ turns a long sum of small-angle arctangents into just the first and last terms — a huge simplification once the denominators are recognized as $n^2+n+1$.

    Step-by-Step Solution

    1. Check the denominators: for $n=1,2,3,4,5$, $n^2+n+1=3,7,13,21,31$ — exactly matching the given series.
    2. Use $\tan^{-1}(n+1)-\tan^{-1}n=\tan^{-1}\dfrac{(n+1)-n}{1+n(n+1)}=\tan^{-1}\dfrac{1}{n^2+n+1}$.
    3. So $\theta=\displaystyle\sum_{n=1}^{5}\left[\tan^{-1}(n+1)-\tan^{-1}n\right]$, which telescopes: all middle terms cancel, leaving $\theta=\tan^{-1}6-\tan^{-1}1$.
    4. $\tan^{-1}1=\pi/4$, so $\theta=\tan^{-1}6-\pi/4$. …

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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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