The number of solutions of tan⁻¹ 4x + tan⁻¹ 6x = π/6, where -1/(2√6)

Q: Why do we have to check the domain? Can't we just solve the quadratic and be done with it?

In JEE Advanced, the domain is the soul of the function. If you ignore it, you are solving for a different function entirely. The identity $\tan^{-1} A + \tan^{-1} B$ has a branch cut. If $AB > 1$, the formula changes to $\pi + \tan^{-1}(\dots)$. By ignoring the domain, you risk including 'ghost' solutions that do not satisfy the original equation.

Q: What happens if $AB > 1$? Does the formula change?

Yes, absolutely! If $AB > 1$, the sum of the angles exceeds the principal range of the inverse tangent function. You would need to add $\pi$ to the result. Always check the product $AB$ first—it is the difference between a correct answer and a trap.

Q: Is there a graphical way to see this?

Yes! Imagine plotting $f(x) = \tan^{-1} 4x + \tan^{-1} 6x$. It is a strictly increasing function. The line $y = \frac{\pi}{6}$ is a horizontal line. They will intersect at exactly one point. The domain constraint effectively 'crops' the graph, ensuring we only count the intersection that falls within the valid window.

Question

The number of solutions of $	an^{-1} 4x + 	an^{-1} 6x = \frac{\pi}{6}$, where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$, is equal to

Options:

(A) 3

(B) 0

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### The Gatekeeper of Trigonometry: A Journey into Inverse Functions

Welcome, future engineer. Today, we are not just solving an equation; we are navigating a minefield. Inverse trigonometry is beautiful, elegant, and notoriously deceptive.

The problem before us, $	an^{-1} 4x + 	an^{-1} 6x = \frac{\pi}{6}$, looks like a standard algebraic exercise. But in the world of JEE Advanced, nothing is ever just 'standard.'

### Phase 1: Respecting the Domain

Before we touch a single variable, look at the constraint: $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$. Why is this here?

In many problems, you might be tempted to ignore such details, treating them as mere 'fine print.' Do not make that mistake. This domain is the gatekeeper.

It ensures that the identity we are about to use is mathematically valid. Without this constraint, our algebraic manipulations could lead us into a realm where the identity fails, creating 'ghost' solutions that do not actually satisfy the original equation.

### Phase 2: The Conditional Identity

We reach for our most powerful tool: the addition identity for inverse tangents:

$$	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$$

But wait! Stop and breathe. This identity is a conditional friend. It is only valid when $AB < 1$.

If $AB > 1$, the sum of the angles shifts, and the formula requires an adjustment of $\pi$. Let us verify our condition. With $A = 4x$ and $B = 6x$, the product is $24x^2$.

Given our domain, $x^2 < \frac{1}{24}$, which implies $24x^2 < 1$. The condition holds! We are safe to proceed. The path is clear.

### Phase 3: The Algebraic Battle

Now, we substitute our values into the identity:

$$	an^{-1} \left( \frac{4x + 6x}{1 - (4x)(6x)} ight) = \frac{\pi}{6}$$

Simplifying this, we get:

$$	an^{-1} \left( \frac{10x}{1 - 24x^2} ight) = \frac{\pi}{6}$$

To free our variable $x$, we apply the tangent function to both sides. Since $	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, we arrive at:

$$\frac{10x}{1 - 24x^2} = \frac{1}{\sqrt{3}}$$

Cross-multiplying gives us a beautiful quadratic equation:

$$10\sqrt{3}x = 1 - 24x^2 \Rightarrow 24x^2 + 10\sqrt{3}x - 1 = 0$$

Do not let the $\sqrt{3}$ intimidate you. It is just a coefficient. We use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)}$$

Calculating the discriminant, $D = 300 + 96 = 396$. Since $\sqrt{396} = 6\sqrt{11}$, our roots are:

$$x = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}$$

### Phase 4: The Final Verdict

We have two potential candidates for $x$. But remember the gatekeeper? We must check if these roots fall within our domain $(-\frac{1}{2\sqrt{6}}, \frac{1}{2\sqrt{6}})$, which is approximately $(-0.204, 0.204)$.

Calculating the values:

$$x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} \approx 0.054$$

$$x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} \approx -0.775$$

Only $x_1$ sits comfortably within our domain. The root $x_2$ is a mathematical artifact—a solution to the quadratic, but not to the original inverse trigonometric equation. We reject it.

Thus, we are left with exactly **1** valid solution.

### Conclusion

This problem is a masterclass in discipline. It teaches us that in mathematics, as in life, the constraints are just as important as the actions.

You navigated the identity, you conquered the quadratic, and you respected the domain. That is how you win at JEE Advanced. Keep this rigor, keep this curiosity, and keep pushing forward.

Animated Solution for Mathematics - Inverse Trigonometric Functions: The number of solutions of $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{6}$ , where $- \frac{1}{2 6} < x < \frac{1}{2 6}$ , is equal to

Select Answer:

Visualized Solution

Understanding the Equation and Domain

The $tan^{- 1}$ Addition Identity

Verifying the Condition $A B < 1$

Applying the Identity

Removing the Inverse Tangent

Forming the Quadratic Equation

Applying the Quadratic Formula

Simplifying the Discriminant

Simplifying the Roots

Evaluating the Numerical Values

Checking the Domain Constraint

Final Conclusion

The Sigma Insight: Solving Inverse Trigonometric Equations

The Gatekeeper of Trigonometry

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Animated Solution for Mathematics - Inverse Trigonometric Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of tan−14x+tan−16x=6π​, where −26​1​<x<26​1​, is equal to