For n ∈ N, if cot⁻¹ 3 + cot⁻¹ 4 + cot⁻¹ 5 + cot⁻¹ n = π/4, then n is equal to

Q: Why can't we just use the $\cot^{-1} x + \cot^{-1} y$ formula?

While it exists, it is rarely taught and prone to errors. Converting to $\tan^{-1}$ is the standard, robust JEE approach that keeps your calculations clean and manageable.

Q: What happens if $xy > 1$?

The formula becomes $\pi + \tan^{-1} (\frac{x+y}{1-xy})$. Always check the product $xy$ to ensure you are in the principal domain. In this problem, the products are always less than $1$, so we are safe!

Q: Is there a faster way?

You could use $\tan(A+B+C+D) = 1$, but that leads to a complex algebraic expansion. The step-by-step reduction is safer and less prone to calculation errors, which is crucial in a high-stakes exam.

Question

For $n \in \mathbb{N}$, if $\cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4}$, then $n$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex equation: 
$$\cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4}$$

It looks intimidating, but mathematics is not about memorizing formulas; it is about finding the right tools to simplify the chaos.

### The Master Key: Conversion

The first step in our journey is to recognize that $\cot^{-1}$ is not our best friend when it comes to addition. Its addition formulas are clunky and rarely used.

Instead, we reach for our most reliable tool: the conversion identity $\cot^{-1} x = 	an^{-1} (\frac{1}{x})$. By applying this to each term, our equation transforms into:

$$	an^{-1} \left( \frac{1}{3} ight) + 	an^{-1} \left( \frac{1}{4} ight) + 	an^{-1} \left( \frac{1}{5} ight) + 	an^{-1} \left( \frac{1}{n} ight) = \frac{\pi}{4}$$

Suddenly, the problem feels much more approachable. We are now working with the familiar territory of tangent addition.

### The Iterative Process: Building Blocks

Now, we must combine these terms using the addition formula:
$$	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$$

Let us take the first two terms: $	an^{-1} (\frac{1}{3}) + 	an^{-1} (\frac{1}{4})$. Here, $x = \frac{1}{3}$ and $y = \frac{1}{4}$.

Since their product $\frac{1}{12}$ is less than $1$, we can proceed:

$$	an^{-1} \left( \frac{\frac{1}{3} + \frac{1}{4}}{1 - \frac{1}{12}} ight) = 	an^{-1} \left( \frac{\frac{7}{12}}{\frac{11}{12}} ight) = 	an^{-1} \left( \frac{7}{11} ight)$$

We have successfully reduced two terms into one. Now, we add the third term, $	an^{-1} (\frac{1}{5})$, to our result:

$$	an^{-1} \left( \frac{7}{11} ight) + 	an^{-1} \left( \frac{1}{5} ight)$$

Again, the product $\frac{7}{55}$ is less than $1$. Applying the formula, we get:

$$	an^{-1} \left( \frac{\frac{7}{11} + \frac{1}{5}}{1 - \frac{7}{55}} ight) = 	an^{-1} \left( \frac{\frac{46}{55}}{\frac{48}{55}} ight) = 	an^{-1} \left( \frac{46}{48} ight) = 	an^{-1} \left( \frac{23}{24} ight)$$

### The Final Pivot: The Subtraction Strategy

We are almost there. Our equation now stands as:
$$	an^{-1} \left( \frac{23}{24} ight) + 	an^{-1} \left( \frac{1}{n} ight) = \frac{\pi}{4}$$

To isolate $	an^{-1} (\frac{1}{n})$, we move the known term to the right:
$$	an^{-1} \left( \frac{1}{n} ight) = \frac{\pi}{4} - 	an^{-1} \left( \frac{23}{24} ight)$$

Remember that $\frac{\pi}{4} = 	an^{-1} (1)$. So, we have:
$$	an^{-1} \left( \frac{1}{n} ight) = 	an^{-1} (1) - 	an^{-1} \left( \frac{23}{24} ight)$$

We use the subtraction formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} (\frac{x-y}{1+xy})$:

$$	an^{-1} \left( \frac{1}{n} ight) = 	an^{-1} \left( \frac{1 - \frac{23}{24}}{1 + 1 \cdot \frac{23}{24}} ight) = 	an^{-1} \left( \frac{\frac{1}{24}}{\frac{47}{24}} ight) = 	an^{-1} \left( \frac{1}{47} ight)$$

Comparing both sides, we find that $\frac{1}{n} = \frac{1}{47}$, which means **$n = 47$**.

And there it is! A systematic, elegant path to the solution. Never fear the complexity of a problem; just break it down, one step at a time.

Animated Solution for Mathematics - Inverse Trigonometric Functions: For $n \in N$ , if $cot^{- 1} 3 + cot^{- 1} 4 + cot^{- 1} 5 + cot^{- 1} n = \frac{π}{4}$ , then $n$ is equal to

Enter Numerical Value:

Visualized Solution

The Given Equation

The Conversion Tool

Rewriting the Equation

The Addition Formula

Combining the First Two Terms

Simplifying the First Result

Adding the Third Term

Simplifying the Second Result

Isolating the Unknown

Using the Tangent of Pi by Four

Final Calculation

The Final Answer

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Master Key

The Iterative Process

The Final Pivot

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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; } For n∈N, if cot−13+cot−14+cot−15+cot−1n=4π​, then n is equal to