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

Solution

Maths sir · 4 steps

Turn every cot⁻¹ into tan⁻¹ of the reciprocal, then add them two at a time.

1. Switch to tan. For x > 0, cot⁻¹ x = tan⁻¹ 1/x. So the equation becomes tan⁻¹ 1/3 + tan⁻¹ 1/4 + tan⁻¹ 1/5 + tan⁻¹ 1/n = π/4. Every product of two of these fractions is less than 1, so the addition formula below applies at each step.

   tan⁻¹ a + tan⁻¹ b = tan⁻¹(a+b)/(1-ab), ab < 1

2. Add the first three. tan⁻¹ 1/3 + tan⁻¹ 1/4 = tan⁻¹(7/12)/(11/12) = tan⁻¹ 7/11. Then tan⁻¹ 7/11 + tan⁻¹ 1/5 = tan⁻¹(46/55)/(48/55) = tan⁻¹ 23/24.
3. Isolate the last term. Now tan⁻¹ 1/n = tan⁻¹ 1 - tan⁻¹ 23/24 = tan⁻¹(1 - 23/24)/(1 + 23/24) = tan⁻¹(1/24)/(47/24) = tan⁻¹ 1/47.
4. So. 1/n = 1/47, which gives n = 47.

The idea to keep. Sums of cot⁻¹ terms are easiest as tan⁻¹ of reciprocals. Combine them in pairs with (a+b)/(1-ab), and use π/4 = tan⁻¹ 1 to pull out the unknown.

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