If α>β>γ>0, then the expression cot ⁻¹{β+(1+β²)/(α-β)}+cot ⁻¹{γ+(1+γ²)/(β-γ)}+cot ⁻¹{α+(1+α²)/(γ-α)} is equal to :

Solution

Maths sir · 4 steps

Turn each term into a difference of tan⁻¹, but the third argument is negative, so add π.

1. One fraction. In the first bracket, β + (1+β²)/(α-β) = (αβ - β² + 1 + β²)/(α-β) = (1+αβ)/(α-β). In the same way the second is (1+βγ)/(β-γ) and the third is (1+γα)/(γ-α).
2. Positive arguments. Since α>β>γ>0, the first two fractions are positive. For x>0, cot⁻¹ x = tan⁻¹ 1/x, so the first term is tan⁻¹(α-β)/(1+αβ) = tan⁻¹ α - tan⁻¹ β. The second is tan⁻¹ β - tan⁻¹ γ.
3. Negative argument. The third fraction is negative because γ<α. For x<0, cot⁻¹ x lies in (π/2, π), so cot⁻¹ x = π + tan⁻¹ 1/x. That makes the third term π + tan⁻¹ γ - tan⁻¹ α.

   (tan⁻¹ α - tan⁻¹ β) + (tan⁻¹ β - tan⁻¹ γ) + (π + tan⁻¹ γ - tan⁻¹ α) = π

4. So. All the tan⁻¹ terms cancel in pairs and only π is left. The expression equals π, option (c).

The idea to keep. cot⁻¹ x = tan⁻¹ 1/x only when x>0. For x<0 add π, because cot⁻¹ always lives in (0, π). Always check the sign of the argument first.

Answer from the NTA final answer key. Solution by PrepWiser.