Q.Prove that $2\sin^{-1} \dfrac{3}{5} = \tan^{-1} \dfrac{24}{7}$.

Showing the 12 most recent of 34 on this concept.

1. 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: $\dfrac{1-x^2}{1+x^2}=\dfrac{24}{25}$.
   5. Cross multiply: $25(1-x^2)=24(1+x^2) \Rightarrow 25-25x^2=24+24x^2 \Rightarrow 1=49x^2 \Rightarrow x^2=\dfrac{1}{49}$.
   6. $x=\pm\dfrac{1}{7}$; matching the positive option given, $x=\dfrac{1}{7}$.

   Common Mistakes

   • Confusing $\sin2\theta$ and $\cos2\theta$ formulas in terms of $\tan\theta$.
   • Sign/arithmetic slip when squaring $3/4$.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{7}$.

   ANSWER: C

2. 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) $\cos\psi=4/5$.
   5. $\sin(2\varphi+\psi)=\sin2\varphi\cos\psi+\cos2\varphi\sin\psi=\dfrac45\cdot\dfrac45+\dfrac35\cdot\dfrac35=\dfrac{16}{25}+\dfrac{9}{25}=\dfrac{25}{25}=1$.

   Common Mistakes

   • Assuming $2\tan^{-1}(1/2)=\tan^{-1}(1)$, which is false — the double-angle formula for $\tan$ must be applied, not simple doubling.
   • Sign errors in $\cos\psi$ (must check which quadrant the inverse-sine value lies in; here it's the principal first-quadrant branch).

   ✓Final answer

   The correct option is (B) 1.

   ANSWER: B

3. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\tan\left(2\,\mathrm{Tan}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Tan}^{-1}\left(\dfrac{1}{7}\right)\right) =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $\sqrt{3}$ (C) $1$ (D) $3/7$

   ›Reveal solutionSolution

   First reduce the double-angle inverse-tangent term to a single tangent value using the tangent double-angle formula, then combine with the second $\tan^{-1}$ term using the tangent addition formula. The result is exactly $1$.

   Concept and Intuition

   Expressions like $\tan(2\tan^{-1}x+\tan^{-1}y)$ are handled in two stages: first collapse $2\tan^{-1}x$ to a single angle whose tangent is known via the double-angle formula $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$, then treat the whole thing as $\tan(\alpha'+\beta)$ using the standard addition formula, where $\alpha'$ is the angle with $\tan\alpha'=\tan(2\tan^{-1}x)$.

   Step-by-Step Solution

   1. Let $\alpha=\tan^{-1}(1/3)$, so $\tan\alpha=1/3$. Then $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac{2}{3}\times\dfrac{9}{8}=\dfrac{18}{24}=\dfrac34$.
   2. Let $\beta=\tan^{-1}(1/7)$, so $\tan\beta=1/7$.
   3. We need $\tan(2\alpha+\beta)=\dfrac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta}=\dfrac{\frac34+\frac17}{1-\frac34\cdot\frac17}$.
   4. Numerator: $\dfrac34+\dfrac17=\dfrac{21+4}{28}=\dfrac{25}{28}$.
   5. Denominator: $1-\dfrac{3}{28}=\dfrac{25}{28}$.
   6. Ratio: $\dfrac{25/28}{25/28}=1$.

   Common Mistakes

   • Trying to add all three angle contributions using a single addition formula at once instead of collapsing $2\tan^{-1}(1/3)$ to its tangent value first.
   • Arithmetic slips when combining the fractions $3/4$ and $1/7$ over a common denominator.

   ✓Final answer

   The correct option is (C) — $1$.

   ANSWER: C

4. 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 $A+B\in(\pi/2,\pi)$.
   4. In this interval, the angle with $\tan=-1$ is $3\pi/4$ (not $-\pi/4$, which lies outside $(\pi/2,\pi)$).
   5. So $A+B=3\pi/4$.

   Common Mistakes

   • Directly writing $\tan^{-1}(-1)=-\pi/4$ as the final answer without checking which branch the true sum lies in.
   • Forgetting that $\tan^{-1}2+\tan^{-1}3$ individually exceed $\pi/4$ each, pushing the sum past $\pi/2$.

   ✓Final answer

   The correct option is (C) — $3\pi/4$.

   ANSWER: C

5. 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. $\phi=\tan^{-1}\!\left(\dfrac{\tan\theta}{3}\right)$.
   7. Then $\sin^{-1}\!\left(\dfrac{3\sin2\theta}{5+4\cos2\theta}\right)=2\phi$ (principal branch), so $\dfrac12\sin^{-1}(\cdots)=\phi=\tan^{-1}\!\left(\dfrac{\tan\theta}{3}\right)$.
   8. Given this equals $\tan^{-1}x$, we get $x=\dfrac{\tan\theta}{3}=\dfrac13\tan\theta$.

   Common Mistakes

   • Trying to directly differentiate or manipulate the inverse-sine expression instead of recognizing the double-angle sine pattern.
   • Choosing the wrong scale factor $u$ (e.g. $u=3t$ instead of $u=t/3$) — check by reconstructing $2u/(1+u^2)$ and matching coefficients carefully.

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{3}\tan\theta$.

   ANSWER: B

6. 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 $\approx5.91°$.
   4. The second term is $\tan^{-1}(1/\sqrt5)=\tan^{-1}(0.4472)\approx24.09°$.
   5. Sum $\approx5.91°+24.09°=30.00°=\pi/6$.

   Common Mistakes

   • Missing the nested-surd simplification and trying to evaluate the messy first term directly (much harder and error-prone).
   • Sign error picking $\sqrt3-\sqrt5$ (negative) instead of $\sqrt5-\sqrt3$.

   ✓Final answer

   The correct option is (A) — $\dfrac{\pi}{6}$.

   ANSWER: A

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. $\sin2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\dfrac{1}{\sqrt{10}}\cdot\dfrac{3}{\sqrt{10}} = \dfrac{6}{10}=\dfrac{3}{5}$.
   3. Let $\beta=\tan^{-1}(2\sqrt2)$: opposite $2\sqrt2$, adjacent $1$, hypotenuse $\sqrt{(2\sqrt2)^2+1^2}=\sqrt{8+1}=3$. So $\cos\beta=\dfrac{1}{3}$.
   4. Sum: $\sin2\alpha+\cos\beta = \dfrac{3}{5}+\dfrac{1}{3} = \dfrac{9}{15}+\dfrac{5}{15}=\dfrac{14}{15}$.

   Common Mistakes

   • Forgetting the factor of 2 in $\sin2\alpha=2\sin\alpha\cos\alpha$.
   • Errors in computing the hypotenuse of the second triangle ($\sqrt{8+1}=3$, not $\sqrt{8}+1$).

   ✓Final answer

   The correct option is (B) — $\dfrac{14}{15}$.

   ANSWER: B

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:

   $\tan^{-1}1-\tan^{-1}\frac19 = \frac{\pi}{4}-\tan^{-1}\frac19.$

   4. Apply the subtraction formula once more: $\tan^{-1}1-\tan^{-1}\frac19 = \tan^{-1}\dfrac{1-\frac19}{1+1\cdot\frac19} = \tan^{-1}\dfrac{8/9}{10/9}=\tan^{-1}\dfrac{8}{10}=\tan^{-1}\dfrac45$.

   Common Mistakes

   • Trying to combine the four arctangents pairwise without spotting the telescoping structure first — this leads to messy nested fractions.
   • Sign errors in the telescoping identity (getting $\tan^{-1}\frac{1}{2n+1}-\tan^{-1}\frac{1}{2n-1}$ backwards), which would make terms add up instead of cancel.

   ✓Final answer

   The correct option is (D) — $\tan^{-1}\left(\dfrac45\right)$.

   ANSWER: D

9. 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 $\theta=\tan^{-1}6-\pi/4$.
   5. $\tan\theta=\tan(\tan^{-1}6-\pi/4)=\dfrac{6-\tan(\pi/4)}{1+6\tan(\pi/4)}=\dfrac{6-1}{1+6}=\dfrac57$.

   Common Mistakes

   • Trying to add the five arctangents directly with the multi-angle tangent-addition formula (extremely messy) instead of spotting the telescoping pattern.
   • Sign error in the final subtraction formula $\tan(x-y)=\dfrac{\tan x-\tan y}{1+\tan x\tan y}$.

   ✓Final answer

   The correct option is (C) — $\dfrac{5}{7}$.

   ANSWER: C

10. 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 $\tan^{-1}(-1)=-\frac{\pi}{4}$.
    4. So $\tan^{-1}2+\tan^{-1}3 = \pi-\frac{\pi}{4}=\frac{3\pi}{4}$ (consistent with the numeric check: $\tan^{-1}2\approx63.43^\circ$, $\tan^{-1}3\approx71.57^\circ$, sum $\approx135^\circ=\frac{3\pi}{4}$).
    5. Therefore the original expression $= -\frac{3\pi}{4}$.

    Common Mistakes

    • Applying the plain formula $\tan^{-1}\frac{a+b}{1-ab}$ without the $\pi$-correction when $ab>1$, which would wrongly give $-\frac{\pi}{4}$ instead of $\frac{3\pi}{4}$.
    • Sign errors handling $\tan^{-1}(-2)$.

    ✓Final answer

    The correct option is (D) — $\dfrac{-3\pi}{4}$.

    ANSWER: D

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.

    D. Let $\theta=\sec^{-1}\dfrac1x$, so $\sec\theta=\dfrac1x \Rightarrow \cos\theta=x \Rightarrow \sin\theta=\sqrt{1-x^2} \Rightarrow \tan\theta=\dfrac{\sqrt{1-x^2}}{x}$. Also $\sin(\tan^{-1}2)=\dfrac{2}{\sqrt5}$ (opposite 2, hypotenuse $\sqrt5$). Equation: $\dfrac{\sqrt{1-x^2}}{x}=\dfrac{2}{\sqrt5}$. Square: $\dfrac{1-x^2}{x^2}=\dfrac45 \Rightarrow 5-5x^2=4x^2 \Rightarrow x^2=\dfrac59 \Rightarrow x=\dfrac{\sqrt5}{3}$ (taking $x>0$). Matches I.

    So: A-II, B-III, C-IV, D-I.

    Common Mistakes

    • In C, forgetting reciprocal tangents summing to $\pi/2$ and instead trying to numerically add the two angles.
    • In D, forgetting to take the positive root given $x>0$.

    ✓Final answer

    The correct option is (B) — A-II, B-III, C-IV, D-I.

    ANSWER: B

12. 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 $=\frac{8281}{8450}$.
    4. Then add $\tan^{-1}\frac1{99}$: combine $\tan^{-1}\frac{8281}{8450}+\tan^{-1}\frac1{99}$ using $\tan^{-1}p+\tan^{-1}q=\tan^{-1}\frac{p+q}{1-pq}$; carrying this through (or simply verifying numerically, as below) confirms the total is exactly $\pi/4$.
    5. Numeric check: $4\tan^{-1}(0.2)=0.789582$, $\tan^{-1}(1/70)=0.014285$, $\tan^{-1}(1/99)=0.010101$. Then $0.789582-0.014285+0.010101=0.785398=\pi/4$ (to 6 places) — confirming the identity.

    Common Mistakes

    • Sign errors in the addition/subtraction formula for $\tan^{-1}$.
    • Not tracking which quadrant the combined angle lies in when the intermediate tangent value exceeds convenient bounds.

    ✓Final answer

    The correct option is (C) — $\dfrac\pi4$.

    ANSWER: C