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

Showing the 12 most recent of 39 on this concept.

1. CBSE 2020Set 65/2/11 markMCQ
   Q.$\tan^{-1} 3 + \tan^{-1}\lambda = \tan^{-1}\left(\dfrac{3+\lambda}{1-3\lambda}\right)$ is valid for what values of $\lambda$? (A) $\lambda \in \left(-\dfrac{1}{3},\ \dfrac{1}{3}\right)$ (B) $\lambda > \dfrac{1}{3}$ (C) $\lambda < \dfrac{1}{3}$ (D) All real values of $\lambda$

   ›Reveal solutionSolution

   The inverse tangent addition formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$ holds only when $xy < 1$. Here $x=3$, $y=\lambda$, so the condition is $3\lambda < 1$, i.e. $\lambda < \frac{1}{3}$. The correct option is (C).

   The formula you’ve written —

   $\tan^{-1} 3 + \tan^{-1}\lambda = \tan^{-1}\left(\frac{3+\lambda}{1-3\lambda}\right)$

   — is the standard inverse tangent addition identity, but it comes with a hidden condition. Many students apply it blindly, and that’s where mistakes happen.

   Let’s understand why the condition exists.

   ---

   The core idea: the range of $\tan^{-1}$ and the product condition

   Recall that $\tan^{-1} x$ (also written $\arctan x$) gives an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$. So the sum of two such angles, $\tan^{-1}3 + \tan^{-1}\lambda$, lies in $(-\pi, \pi)$.

   The formula

   $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$

   is derived from the tangent addition formula:

   $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

   If we set $A = \tan^{-1}x$, $B = \tan^{-1}y$, then $\tan(A+B) = \frac{x+y}{1-xy}$.

   But here’s the catch: $\tan^{-1}\frac{x+y}{1-xy}$ always gives an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$. So the equality holds only when $A+B$ itself lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$.

   When does $A+B$ stay inside that interval? It turns out the cleanest condition is $xy < 1$.

   $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy} \quad \text{if and only if} \quad xy < 1$

   If $xy = 1$, the denominator is zero — the formula breaks. If $xy > 1$, then $A+B$ falls outside $(-\frac{\pi}{2}, \frac{\pi}{2})$, and the right-hand side would give a different principal value (you’d need to add or subtract $\pi$).

   ---

   Applying it to this problem

   Here $x = 3$ and $y = \lambda$. So the condition for the formula to be valid is:

   1. Write the product condition:

      $xy < 1 \quad\Rightarrow\quad 3\lambda < 1$

   2. Solve for $\lambda$:

      $\lambda < \frac{1}{3}$

   That’s it. No further restrictions — $\lambda$ can be any real number less than $\frac{1}{3}$.

   Watch out

   A common mistake is to also consider the denominator $1-3\lambda \neq 0$, i.e. $\lambda \neq \frac{1}{3}$. But that’s already covered: $3\lambda < 1$ excludes $\lambda = \frac{1}{3}$ automatically. The real pitfall is forgetting the product condition entirely and assuming the formula works for all $\lambda$.

   ---

   Matching with the options

   • (A) $\lambda \in \left(-\frac{1}{3}, \frac{1}{3}\right)$ — too restrictive; $\lambda$ can be much smaller than $-\frac{1}{3}$ and the formula still holds.
   • (B) $\lambda > \frac{1}{3}$ — exactly the opposite of the condition.
   • (C) $\lambda < \frac{1}{3}$ — correct.
   • (D) All real values — false, as shown.

   Tip

   Quick check: try $\lambda = 0$. Then LHS = $\tan^{-1}3 + 0$, RHS = $\tan^{-1}3$ — works. Try $\lambda = 1$. Then LHS = $\tan^{-1}3 + \frac{\pi}{4} \approx 1.249 + 0.785 = 2.034$ rad, but RHS = $\tan^{-1}\frac{4}{-2} = \tan^{-1}(-2) \approx -1.107$ rad — clearly not equal. So the condition is real.

   ✓Final answer

   The formula is valid for $\lambda < \frac{1}{3}$, which corresponds to option (C).

2. 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 $\tan\theta = -\frac{1}{\sqrt{3}}$.

   Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\tan$ is odd, $\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$.

   And $-\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so it is the principal value.

   ✓Final answer

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

3. 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 $x = \frac{1}{3}$:

   $\frac{2\cdot\frac{1}{3}}{1-\frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3}\cdot\frac{9}{8} = \frac{18}{24} = \frac{3}{4}.$

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

   ✓Final answer

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

4. 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

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

   ✓Final answer

   (c) $0$.

5. 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

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

   ✓Final answer

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

6. 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:

   $1+ab = 1 + \frac{x}{y}\cdot\frac{x-y}{x+y} = \frac{y(x+y)+x(x-y)}{y(x+y)} = \frac{xy+y^2+x^2-xy}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

   So the ratio is $1$, giving

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

   ✓Final answer

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

7. 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

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

   ✓Final answer

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

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

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

   Since $\frac{5\pi}{14}$ lies in the principal range $(0,\pi)$ of $\cot^{-1}$,

   $\cot^{-1}\left(\tan\frac{\pi}{7}\right)=\cot^{-1}\left(\cot\frac{5\pi}{14}\right)=\frac{5\pi}{14}.$

   ✓Final answer

   (B) $\frac{5\pi}{14}$.

9. 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 $\tan^{-1}(-x)=-\tan^{-1}x$,

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

   This lies in the principal range, so it is the required value.

   ✓Final answer

   (D) $-\frac{\pi}{3}$.

10. 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 $\cot\frac{\pi}{6}=\sqrt{3}$,

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

    Therefore

    $\tan^{-1}(\sqrt{3})-\cot^{-1}(-\sqrt{3})=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}.$

    ✓Final answer

    (B) $-\frac{\pi}{2}$.

11. 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

    $\tan^{-1}\frac12+\tan^{-1}\frac13=\tan^{-1}\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}}=\tan^{-1}1=\frac{\pi}{4}.$

    ✓Final answer

    (B) $\frac{\pi}{4}$.

12. 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:

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

    Hence

    $\tan\left(\frac{\pi}{4}\right) = 1.$

    ✓Final answer

    (A) $1$.