Q.Differentiate $\tan^{-1}\left(\dfrac{1 + \cos x}{\sin x}\right)$ with respect to $x$.

Showing the 12 most recent of 23 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If $2\cot^{-1}\left(\dfrac{4}{3}\right) = \cos^{-1}\left(\dfrac{x}{5}\right)$ , then the value of $x$ is equal to (A) $\dfrac{3}{25}$ (B) $\dfrac{7}{25}$ (C) $\dfrac{3}{5}$ (D) $\dfrac{7}{5}$ (E) $\dfrac{5}{7}$

   ›Reveal solutionSolution

   Compute $\cos(2\cot^{-1}\tfrac43)=\tfrac{7}{25}$, so $x/5=7/25$ gives $x=\tfrac75$.

   Let $\phi=\cot^{-1}\dfrac43$, so $\cot\phi=\dfrac43$, giving a $3\text{-}4\text{-}5$ triangle with $\sin\phi=\dfrac35$, $\cos\phi=\dfrac45$. Then

   $\cos2\phi=1-2\sin^2\phi=1-2\cdot\frac{9}{25}=\frac{7}{25}.$ …

   🔒 Start your 14-day free trial to see the full solution →
2. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\tan\left(\tan^{-1}(3)+\tan^{-1}(7)\right)$ is equal to (A) $-\dfrac{1}{2}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{5}$ (D) $-\dfrac{1}{5}$ (E) $0$

   ›Reveal solutionSolution

   Use $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{10}{-20}=-\tfrac12$.

   With $\tan A=3$, $\tan B=7$: …

   🔒 Start your 14-day free trial to see the full solution →
3. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.If $tan\,\alpha = \frac{5}{6}$ and $tan\,\beta = \frac{1}{11}$, where $0 < \alpha, \beta < \frac{\pi}{2}$ then $\alpha + \beta =$ (A) $\frac{\pi}{6}$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{4}$ (E) $\frac{2\pi}{3}$

   ›Reveal solutionSolution

   Use the addition formula for tangent: $\tan(\alpha+\beta)=1$, hence $\alpha+\beta=\tfrac{\pi}{4}$.

   $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{\tfrac56+\tfrac1{11}}{1-\tfrac{5}{66}}=\frac{\tfrac{55+6}{66}}{\tfrac{66-5}{66}}=\frac{61}{61}=1.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.Let $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\sin x - \cos x}\right)$, $0 < x < \frac{\pi}{2}$. Then $\frac{dy}{dx}$ is equal to (A) $x$ (B) $-1$ (C) $-x$ (D) $2x$ (E) $-2x$

   ›Reveal solutionSolution

   Rewrite the fraction using $\sin x\pm\cos x=\sqrt2\,(\ldots)$ to collapse the arctan into a linear function.

   $\sin x+\cos x=\sqrt{2}\cos\left(x-\dfrac{\pi}{4}\right)$ and $\sin x-\cos x=\sqrt{2}\sin\left(x-\dfrac{\pi}{4}\right)$.

   So the ratio $=\cot\left(x-\dfrac{\pi}{4}\right)=\tan\left(\dfrac{\pi}{2}-\left(x-\dfrac{\pi}{4}\right)\right)=\tan\left(\dfrac{3\pi}{4}-x\right)$. …

   🔒 Start your 14-day free trial to see the full solution →
5. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.The value of $2\tan^{-1}\left(\frac{1}{3}\right)+\cot^{-1}\left(\frac{3}{4}\right)=$ (A) $\frac{\pi}{3}$ (B) $\frac{2\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{6}$ (E) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   $2\tan^{-1}\frac13=\tan^{-1}\frac34$; adding $\tan^{-1}\frac43$ (its complementary reciprocal) gives $\frac{\pi}{2}$.

   Using the double-angle formula:

   $2\tan^{-1}\frac13=\tan^{-1}\!\frac{2\cdot\frac13}{1-\frac19}=\tan^{-1}\!\frac{\frac23}{\frac89}=\tan^{-1}\frac34.$ …

   🔒 Start your 14-day free trial to see the full solution →
6. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The value of $\tan^{-1}\left(\frac{\cos x - \sqrt{3}\sin x}{\sqrt{3}\cos x + \sin x}\right)$, where $0 < x < \frac{\pi}{2}$ is (A) $\frac{\pi}{6} - x$ (B) $\frac{\pi}{4} - x$ (C) $\frac{\pi}{3} - x$ (D) $\frac{\pi}{2} - x$ (E) $\pi - x$

   ›Reveal solutionSolution

   Recognize numerator and denominator as $2\cos$ and $2\sin$ of $\left(x+\frac{\pi}{3}\right)$.

   Numerator: $\cos x-\sqrt3\sin x=2\left(\tfrac12\cos x-\tfrac{\sqrt3}{2}\sin x\right)=2\cos\!\left(x+\tfrac{\pi}{3}\right)$.

   Denominator: $\sqrt3\cos x+\sin x=2\left(\tfrac{\sqrt3}{2}\cos x+\tfrac12\sin x\right)=2\sin\!\left(x+\tfrac{\pi}{3}\right)$. …

   🔒 Start your 14-day free trial to see the full solution →
7. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.If $f(x) = \tan^{-1}\left(\frac{3\cos x - 5\sin x}{5\cos x + 3\sin x}\right)$, then the value $f'(1)$ is (A) $1$ (B) $\frac{1}{2}$ (C) $\frac{1}{4}$ (D) $-1$ (E) $-\frac{1}{4}$

   ›Reveal solutionSolution

   Reduce the argument to a $\tan^{-1}(A)-x$ form so the derivative is a constant.

   Divide numerator and denominator by $5\cos x$:

   $\frac{3\cos x-5\sin x}{5\cos x+3\sin x}=\frac{\frac35-\tan x}{1+\frac35\tan x}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.If $f(x) = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $f\left(\dfrac{1}{\sqrt{3}}\right)$ is equal to (A) $\dfrac{\pi}{6}$ (B) $\dfrac{2\pi}{3}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{4\pi}{3}$ (E) $0$

   ›Reveal solutionSolution

   Using tan^-1(2x/(1-x^2)) = 2 tan^-1 x for |x|<1, f(1/sqrt3) = 2*(pi/6) = pi/3.

   Concept and Intuition

   The identity 2 tan^-1 x = tan^-1(2x/(1-x^2)) holds for |x| < 1. Since 1/sqrt3 < 1, the formula applies directly.

   Step-by-Step Solution

   1. f(x) = tan^-1(2x/(1-x^2)) = 2 tan^-1 x for |x| < 1.
   2. tan^-1(1/sqrt3) = pi/6.
   3. f(1/sqrt3) = 2 * pi/6 = pi/3. …

   🔒 Start your 14-day free trial to see the full solution →
9. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$cot^{-1}(1) + cot^{-1}(2) + cot^{-1}(3) =$ (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $\frac{3\pi}{2}$ (D) $\pi$ (E) $0$

   ›Reveal solutionSolution

   Convert to $\tan^{-1}$ and use the addition formula.

   $\cot^{-1}1 = \tfrac{\pi}{4}$. For the other two, $\cot^{-1}2 + \cot^{-1}3 = \tan^{-1}\tfrac12 + \tan^{-1}\tfrac13$: …

   🔒 Start your 14-day free trial to see the full solution →
10. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.If $\tan^{-1}x = \tan^{-1}(3) - \dfrac{\pi}{4}$, then $x$ is equal to (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{4}$ (C) $1$ (D) $3$ (E) $2$

    ›Reveal solutionSolution

    Write $\frac{\pi}{4}=\tan^{-1}1$ and use the subtraction formula: $\tan^{-1}3-\tan^{-1}1=\tan^{-1}\frac12\Rightarrow x=\frac12$.

    Apply the identity. $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ with $a=3,b=1$: …

    🔒 Start your 14-day free trial to see the full solution →
11. KEAM 2025Set eng-2025-04274 marksMCQ
    Q.$\tan^{-1}\left(\dfrac{1001}{999}\right)-\tan^{-1}\left(\dfrac{2}{2000}\right) =$ (A) $\dfrac{\pi}{3}$ (B) $\pi$ (C) $1$ (D) $\dfrac{\pi}{6}$ (E) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    Rewrite $\dfrac{2}{2000}=\dfrac{1}{1000}$ and use the arctangent subtraction formula; the numerator and denominator turn out equal, giving $\tan^{-1}(1)=\dfrac{\pi}{4}$.

    We evaluate $\tan^{-1}\!\left(\dfrac{1001}{999}\right)-\tan^{-1}\!\left(\dfrac{2}{2000}\right)$.

    Note $\dfrac{2}{2000}=\dfrac{1}{1000}$. Using $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}$ with $a=\dfrac{1001}{999}$, $b=\dfrac{1}{1000}$: …

    🔒 Start your 14-day free trial to see the full solution →
12. KEAM 2025Set eng-2025-04294 marksMCQ
    Q.If $\tan^{-1}2x + \tan^{-1}3x = \dfrac{\pi}{4}$, then the value of $x$ is equal to (A) $\dfrac{1}{6}$ (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{2}$ (E) $1$

    ›Reveal solutionSolution

    $\tan^{-1}2x+\tan^{-1}3x=\frac\pi4$ gives $\frac{5x}{1-6x^2}=1$, i.e. $6x^2+5x-1=0$; roots $x=\frac16,-1$, and only $x=\frac16$ satisfies the equation. …

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