If α>β>γ>0, then the expression cot−1{β+(α−β)(1+β5)}+cot−1{γ+(β−γ)(1+γ2)}+cot−1{α+(γ−α)(1+α2)} is equal to:
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Step-by-Step Solution
Step 1: Recall the inverse cotangent identity
We will use the inverse cotangent identity to simplify the given expression. Specifically, the identity cot−1x−cot−1y=cot−1(y−xxy+1) 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 β+(α−β)(1+β5). This does not directly match the form y−xxy+1. However, if we consider the identity cot−1x−cot−1y=cot−1(y−xxy+1), and let x=α and y=β, then y−x=β−α. The given term is cot−1(α−βαβ+1). This can be rewritten as cot−1α−cot−1β if we assume the form y−xxy+1 where x=α and y=β. However, the problem statement has a typo in the exponents. Assuming the terms are meant to be of the form cot−1(y−xxy+1), we can rewrite the first term as cot−1α−cot−1β. The given expression has β5 and γ2 which are likely typos and should be β and γ respectively to fit the pattern. Assuming the intended form is cot−1(α−βαβ+1), this simplifies to cot−1α−cot−1β.
Step 3: Rewrite the second term
Similarly, for the second term, assuming the intended form is cot−1(β−γβγ+1), it simplifies to cot−1β−cot−1γ. This follows the same pattern as the first term, where the variables are β and γ.
Step 4: Rewrite the third term
And for the third term, assuming the intended form is cot−1(γ−αγα+1), it simplifies to cot−1γ−cot−1α. 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−1β cancels with +cot−1β, and −cot−1γ cancels with +cot−1γ, and +cot−1α cancels with −cot−1α. This leaves us with a final sum of 0.