Q.Prove that $\mathrm{Tan}^{-1}\dfrac{1}{2} + \mathrm{Tan}^{-1}\dfrac{1}{5} + \mathrm{Tan}^{-1}\dfrac{1}{8} = \dfrac{\pi}{4}$.

Showing the 12 most recent of 39 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) =$

   (a) $\frac{\pi}{3}$

   (b) $\frac{\pi}{6}$

   (c) $-\frac{\pi}{3}$

   (d) $-\frac{\pi}{6}$

   ›Reveal solutionSolution

   $\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$.

   The principal value of $\tan^{-1}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

   We need the angle $\theta$ in this range with $\tan\theta = -\frac{1}{\sqrt{3}}$.

   …

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2. CBSE 2026Set A1 markMCQ
   Q.$2\tan^{-1}\frac{1}{3} =$

   (a) $\tan^{-1}\frac{3}{2}$

   (b) $\tan^{-1}\frac{3}{4}$

   (c) $\tan^{-1}\frac{4}{3}$

   (d) $\tan^{-1}\frac{2}{3}$

   ›Reveal solutionSolution

   $2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$.

   Use the double-angle identity valid for $|x|<1$:

   $2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}.$

   Here $x = \frac{1}{3}$: …

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3. CBSE 2026Set A1 markMCQ
   Q.$x \in R,\ \cot(\tan^{-1}x + \cot^{-1}x) =$

   (a) $1$

   (b) $\frac{1}{2}$

   (c) $0$

   (d) $\frac{1}{3}$

   ›Reveal solutionSolution

   $\cot\left(\tan^{-1}x + \cot^{-1}x\right) = \cot\frac{\pi}{2} = 0$.

   For all real $x$, the complementary identity gives

   $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.$

   Therefore …

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4. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}2 + \tan^{-1}3 =$

   (a) $-\frac{\pi}{4}$

   (b) $\frac{\pi}{4}$

   (c) $\frac{3\pi}{4}$

   (d) $\pi$

   ›Reveal solutionSolution

   $\tan^{-1}2 + \tan^{-1}3 = \frac{3\pi}{4}$.

   Use the addition formula. With $a=2,\ b=3$ we have $ab=6>1$, so

   $\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\frac{a+b}{1-ab}.$

   Compute:

   $\frac{a+b}{1-ab} = \frac{5}{1-6} = \frac{5}{-5} = -1.$

   So …

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5. CBSE 2026Set A1 markMCQ
   Q.$\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} =$

   (a) $-\frac{3\pi}{4}$

   (b) $\frac{\pi}{2}$

   (c) $\frac{\pi}{4}$

   (d) $\frac{\pi}{3}$

   ›Reveal solutionSolution

   $\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} = \frac{\pi}{4}$.

   Use $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$ with $a=\frac{x}{y}$, $b=\frac{x-y}{x+y}$.

   Numerator:

   $a-b = \frac{x}{y} - \frac{x-y}{x+y} = \frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2+xy-xy+y^2}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

   Denominator: …

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6. CBSE 2026Set A1 markMCQ
   Q.$|x| \le 1,\ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) =$

   (a) $2\cos^{-1}x$

   (b) $2\sin^{-1}x$

   (c) $2\tan^{-1}x$

   (d) $\tan^{-1}2x$

   ›Reveal solutionSolution

   $\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x$ (for $0\le x\le 1$).

   Put $x = \tan\theta$, so $\theta = \tan^{-1}x$. Then

   $\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta.$

   Therefore …

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7. CBSE 2025Set E1 markMCQ
   Q.$\cot^{-1}\left(\tan\frac{\pi}{7}\right) =$

   (a) $\frac{\pi}{7}$

   (b) $\frac{5\pi}{14}$

   (c) $\frac{9\pi}{14}$

   (d) $\frac{3\pi}{14}$

   ›Reveal solutionSolution

   Convert the tangent to a cotangent using complementary angles; the answer is $\frac{5\pi}{14}$.

   Use $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$:

   $\tan\frac{\pi}{7}=\cot\left(\frac{\pi}{2}-\frac{\pi}{7}\right)=\cot\frac{7\pi-2\pi}{14}=\cot\frac{5\pi}{14}.$

   …

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8. CBSE 2025Set E1 markMCQ
   Q.$\tan^{-1}(-\sqrt{3}) =$

   (a) $\frac{\pi}{6}$

   (b) $\frac{\pi}{3}$

   (c) $\frac{2\pi}{3}$

   (d) $-\frac{\pi}{3}$

   ›Reveal solutionSolution

   $\tan^{-1}$ is an odd function with principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$; the value is $-\frac{\pi}{3}$.

   Since $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan^{-1}(-x)=-\tan^{-1}x$, …

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9. CBSE 2025Set E1 markMCQ
   Q.$\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) =$

   (a) $0$

   (b) $-\frac{\pi}{2}$

   (c) $\pi$

   (d) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   Evaluate each inverse function in its principal range and subtract; result $-\frac{\pi}{2}$.

   First, $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$.

   For $\cot^{-1}(-\sqrt{3})$, the principal range of $\cot^{-1}$ is $(0,\pi)$. We need $\cot\theta=-\sqrt{3}$ with $\theta\in(0,\pi)$. Since $\cot\frac{\pi}{6}=\sqrt{3}$,

   $\cot^{-1}(-\sqrt{3})=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$

   …

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10. CBSE 2025Set E1 markMCQ
    Q.$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} =$

    (a) $\pi$

    (b) $\frac{\pi}{4}$

    (c) $\frac{\pi}{2}$

    (d) $\frac{\pi}{3}$

    ›Reveal solutionSolution

    Use the sum formula for inverse tangents; the sum is $\frac{\pi}{4}$.

    When $xy<1$, $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$. Here $xy=\frac16<1$, so …

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11. CBSE 2025Set E1 markMCQ
    Q.$\tan\left\{\frac{1}{2}\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right)\right\} =$

    (a) $1$

    (b) $\sqrt{3}$

    (c) $0$

    (d) $\infty$

    ›Reveal solutionSolution

    $\tan^{-1}x + \tan^{-1}\frac1x = \frac{\pi}{2}$; half is $\frac{\pi}{4}$; $\tan\frac{\pi}{4} = 1$.

    For $x > 0$ there is a standard identity:

    $\tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2}.$

    Taking half: …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.For what value of $x$, $2\tan^{-1}x=\sin^{-1}\left(\dfrac{\sqrt3}{2}\right)$?

    (i) $\dfrac{1}{\sqrt3}$

    (ii) $\dfrac13$

    (iii) $\dfrac14$

    (iv) $\dfrac{1}{\sqrt5}$

    ›Reveal solutionSolution

    Simplify the RHS first, then solve for $x$.

    $\sin^{-1}\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{3}$

    So the equation becomes: …

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