If , and are in AP and , and are also in AP, then
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
Since x, y, z are in AP: x + z = 2y.
Since tan⁻¹x, tan⁻¹y, tan⁻¹z are in AP: tan⁻¹x + tan⁻¹z = 2tan⁻¹y.
Using the addition formula: tan⁻¹x + tan⁻¹z = tan⁻¹((x+z)/(1-xz)), and 2tan⁻¹y = tan⁻¹(2y/(1-y²)).
So (x+z)/(1-xz) = 2y/(1-y²). Substituting x+z = 2y: 2y/(1-xz) = 2y/(1-y²), which gives 1-xz = 1-y², i.e., xz = y².
But x, y, z in AP also gives xz ≤ y² (AM-GM), with equality only when x = y = z. Since both xz = y² and xz ≤ y² hold simultaneously, we must have x = y = z.
Since tan⁻¹x, tan⁻¹y, tan⁻¹z are in AP: tan⁻¹x + tan⁻¹z = 2tan⁻¹y.
Using the addition formula: tan⁻¹x + tan⁻¹z = tan⁻¹((x+z)/(1-xz)), and 2tan⁻¹y = tan⁻¹(2y/(1-y²)).
So (x+z)/(1-xz) = 2y/(1-y²). Substituting x+z = 2y: 2y/(1-xz) = 2y/(1-y²), which gives 1-xz = 1-y², i.e., xz = y².
But x, y, z in AP also gives xz ≤ y² (AM-GM), with equality only when x = y = z. Since both xz = y² and xz ≤ y² hold simultaneously, we must have x = y = z.
[JEE Main 2013]