The sum of possible values of x for
tan⁻¹(x + 1) + cot⁻¹( 1/(x - 1) ) = tan⁻¹( 8/31 ) is :

Solution

Maths sir · 4 steps

Get the quadratic quickly, but test both roots: cot⁻¹ of a negative number hides an extra π.

1. Rewrite cot inverse. Let y = x-1. Since cot⁻¹ takes values in (0, π), cot⁻¹ 1/y = tan⁻¹ y when y > 0, but cot⁻¹ 1/y = π + tan⁻¹ y when y < 0.
2. Take tan. In both cases the tan of the left side is ((x+1)+(x-1))/(1-(x+1)(x-1)) = 2x/(2-x²). Set this equal to 8/31 and solve.

   2x/(2-x²)=8/31 ⇒ 4x²+31x-8=0 ⇒ (4x-1)(x+8)=0

3. Check the branch. Both roots, x = 1/4 and x = -8, are less than 1, so x - 1 < 0. The equation is then really tan⁻¹(x+1)+tan⁻¹(x-1) = tan⁻¹ 8/31 - π, which is about -2.89.

   Root	tan⁻¹(x+1)+tan⁻¹(x-1)	Equals tan⁻¹ 8/31-π?
   x = 1/4	tan⁻¹ 5/4 - tan⁻¹ 3/4 = tan⁻¹ 8/31, which is positive	No, it is off by π, so reject
   x = -8	-(tan⁻¹ 7+tan⁻¹ 9) = -(π - tan⁻¹ 8/31)	Yes, so accept

4. So. Only x = -8 satisfies the equation, so the sum of possible values is -8 = -32/4, which is option (a).

The idea to keep. Taking tan of both sides loses track of which branch the angle is on. In inverse trig equations, put each root back and check the actual angle, not just its tan.

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