For any y ∈ ℝ, let cot⁻¹(y) ∈ (0, π) and tan⁻¹(y) ∈ (-π/2, π/2). Then the sum of all the solutions o

Q: Wait, why do we reject the negative root in Case 1?

In Case 1, we assumed $y \in (0, 3)$. When we solved the quadratic equation $y^2 + 2\sqrt{3}y - 9 = 0$, we obtained two roots: $\sqrt{3}$ and $-3\sqrt{3}$. Since our initial assumption was that $y$ must be positive, $-3\sqrt{3}$ is invalid. Always check your final roots against the domain constraints of your specific case!

Q: Why can't we just use the identity $\tan^{-1}(X) + \cot^{-1}(1/X) = \pi/2$?

That identity is only true for $X > 0$. When $X < 0$, the relationship changes because the range of $\cot^{-1}$ is $(0, \pi)$. Specifically, for $X < 0$, $\tan^{-1}(X) + \cot^{-1}(1/X) = -\pi/2$. This is a classic JEE trap! Always verify the sign of the argument before applying inverse trig identities.

Q: How do I know the sign of $X = \frac{6y}{9-y^2}$?

Look at the denominator: $9 - y^2$. Since the domain is $0 0$, $X > 0$. If $y < 0$, $X < 0$. It is that simple!

Question

For any $y \in \mathbb{R}$, let $\cot^{-1}(y) \in (0, \pi)$ and $	an^{-1}(y) \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Then the sum of all the solutions of the equation $	an^{-1}\left(\frac{6y}{9-y^2}ight) + \cot^{-1}\left(\frac{9-y^2}{6y}ight) = \frac{2\pi}{3}$ for $0 < |y| < 3$, is equal to

Options:

(A) $2\sqrt{3}-3$

(B) $3-2\sqrt{3}$

(C) $4\sqrt{3}-6$

(D) $6-4\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Beauty of Branching Paths Welcome, fellow traveler on the JEE journey! Today, we are going to dismantle a problem that, at first glance, looks like a tangled mess of inverse trigonometry. But as we peel back the layers, you will see that it is actually a beautiful exercise in domain analysis and logical branching. ### Phase 1: Simplifying the Beast Look at the equation: $$ an^{-1}\left(\frac{6y}{9-y^2} ight) + \cot^{-1}\left(\frac{9-y^2}{6y} ight) = \frac{2\pi}{3}$$ The key to conquering such expressions is to find the hidden symmetry. Notice that the argument of the second term is the exact reciprocal of the first. Let us define a substitution: $$X = \frac{6y}{9-y^2}$$ Now, our equation transforms into the much friendlier: $$ an^{-1}(X) + \cot^{-1}\left(\frac{1}{X} ight) = \frac{2\pi}{3}$$ ### Phase 2: The Domain Trap Before we proceed, we must respect the boundaries. We are given $0 < |y| < 3$. This means $y$ can be in the interval $(0, 3)$ or $(-3, 0)$. Now, look at the denominator of our substitution $X$, which is $9 - y^2$. Since $|y| < 3$, it follows that $y^2 < 9$, meaning $9 - y^2$ is always positive. This is a crucial realization! Because the denominator is always positive, the sign of $X$ is entirely dependent on the sign of $y$. If $y$ is positive, $X$ is positive; if $y$ is negative, $X$ is negative. ### Phase 3: The Branching Path This is where the JEE examiners test your conceptual depth. The identity for $\cot^{-1}(1/X)$ is not a single, universal formula. It branches: 1. If $X > 0$, then $\cot^{-1}(1/X) = an^{-1}(X)$. 2. If $X < 0$, then $\cot^{-1}(1/X) = \pi + an^{-1}(X)$. We must solve this in two distinct cases. **Case 1: $y \in (0, 3)$** Here, $X > 0$. Our equation becomes: $$ an^{-1}(X) + an^{-1}(X) = \frac{2\pi}{3}$$ This simplifies to $2 an^{-1}(X) = \frac{2\pi}{3}$, or $ an^{-1}(X) = \frac{\pi}{3}$. Taking the tangent of both sides, we get $X = an(\pi/3) = \sqrt{3}$. Substituting back, we have: $$\frac{6y}{9-y^2} = \sqrt{3} \implies y^2 + 2\sqrt{3}y - 9 = 0$$ Solving this quadratic, we find $y = \sqrt{3}$ or $y = -3\sqrt{3}$. Since we are in the positive domain, we accept **$y = \sqrt{3}$**. **Case 2: $y \in (-3, 0)$** Here, $X < 0$. Our equation becomes: $$ an^{-1}(X) + (\pi + an^{-1}(X)) = \frac{2\pi}{3}$$ This simplifies to $2 an^{-1}(X) = \frac{2\pi}{3} - \pi = -\frac{\pi}{3}$, so $ an^{-1}(X) = -\frac{\pi}{6}$. Thus, $X = an(-\pi/6) = -\frac{1}{\sqrt{3}}$. Substituting back, we have: $$\frac{6y}{9-y^2} = -\frac{1}{\sqrt{3}} \implies y^2 - 6\sqrt{3}y - 9 = 0$$ The roots are $y = 3\sqrt{3} \pm 6$. Since we need $y \in (-3, 0)$, we accept **$y = 3\sqrt{3} - 6$**. ### Phase 4: The Final Synthesis We have our two valid solutions: $y_1 = \sqrt{3}$ and $y_2 = 3\sqrt{3} - 6$. The problem asks for the sum of all solutions. Adding them together: $$\sqrt{3} + 3\sqrt{3} - 6 = 4\sqrt{3} - 6$$ And there you have it! By carefully analyzing the domain and respecting the branching nature of inverse trigonometric identities, we have turned a complex problem into a series of logical, manageable steps. Keep practicing this level of rigor, and you will master the JEE! The final answer is **$4\sqrt{3} - 6$**.

Select Answer:

Visualized Solution

Introduce Substitution

Analyze Domain and Sign

Inverse Trigonometry Property

Case 1: Positive $y$

Solve for $X$ (Case 1)

Substitute $X$ Back (Case 1)

Solve Quadratic (Case 1)

Filter Roots (Case 1)

Case 2: Negative $y$

Solve for $X$ (Case 2)

Substitute $X$ Back (Case 2)

Solve Quadratic (Case 2)

Filter Roots (Case 2)

Final Sum of Solutions

The Sigma Insight: Properties of Inverse Trigonometric Functions

The Beauty of Branching Paths

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Introduce Substitution

Analyze Domain and Sign

Inverse Trigonometry Property

Case 1: Positive .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X (Case 1)

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X Back (Case 1)