If $\alpha > \beta > \gamma > 0$, then the expression $\cot^{-1} \left\{\beta + \frac{(1 + \beta^5)}{(\alpha - \beta)}\right\} + \cot^{-1} \left\{\gamma + \frac{(1 + \gamma^2)}{(\beta - \gamma)}\right\} + \cot^{-1} \left\{\alpha + \frac{(1 + \alpha^2)}{(\gamma - \alpha)}\right\}$ is equal to:

Step 1: Recall the inverse cotangent identity

We will use the inverse cotangent identity to simplify the given expression. Specifically, the identity $\cot^{-1} x - \cot^{-1} y = \cot^{-1} \left(\frac{xy + 1}{y - x}\right)$ will be useful here. This identity allows us to express the sum or difference of two inverse cotangent functions as a single inverse cotangent function.

Step 2: Rewrite the first term

Let's analyze the first term. The expression inside the $\cot^{-1}$ function is $\beta + \frac{(1 + \beta^5)}{(\alpha - \beta)}$. This does not directly match the form $\frac{xy+1}{y-x}$. However, if we consider the identity $\cot^{-1} x - \cot^{-1} y = \cot^{-1} \left(\frac{xy + 1}{y - x}\right)$, and let $x = \alpha$ and $y = \beta$, then $y-x = \beta - \alpha$. The given term is $\cot^{-1} \left(\frac{\alpha\beta + 1}{\alpha - \beta}\right)$. This can be rewritten as $\cot^{-1} \alpha - \cot^{-1} \beta$ if we assume the form $\frac{xy+1}{y-x}$ where $x=\alpha$ and $y=\beta$. However, the problem statement has a typo in the exponents. Assuming the terms are meant to be of the form $\cot^{-1} \left(\frac{xy+1}{y-x}\right)$, we can rewrite the first term as $\cot^{-1} \alpha - \cot^{-1} \beta$. The given expression has $\beta^5$ and $\gamma^2$ which are likely typos and should be $\beta$ and $\gamma$ respectively to fit the pattern. Assuming the intended form is $\cot^{-1} \left(\frac{\alpha\beta + 1}{\alpha - \beta}\right)$, this simplifies to $\cot^{-1} \alpha - \cot^{-1} \beta$.

Step 3: Rewrite the second term

Similarly, for the second term, assuming the intended form is $\cot^{-1} \left(\frac{\beta\gamma + 1}{\beta - \gamma}\right)$, it simplifies to $\cot^{-1} \beta - \cot^{-1} \gamma$. This follows the same pattern as the first term, where the variables are $\beta$ and $\gamma$.

Step 4: Rewrite the third term

And for the third term, assuming the intended form is $\cot^{-1} \left(\frac{\gamma\alpha + 1}{\gamma - \alpha}\right)$, it simplifies to $\cot^{-1} \gamma - \cot^{-1} \alpha$. This completes the pattern for all three terms.

Step 5: Sum the simplified terms

Now we substitute these simplified forms back into the original expression. We will notice a telescoping sum, where intermediate terms cancel each other out.

Step 6: Calculate the final sum

As we sum the terms, $-\cot^{-1} \beta$ cancels with $+\cot^{-1} \beta$, and $-\cot^{-1} \gamma$ cancels with $+\cot^{-1} \gamma$, and $+\cot^{-1} \alpha$ cancels with $-\cot^{-1} \alpha$. This leaves us with a final sum of 0.

More from this chapter

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