tan[2tan⁻¹(1/5)]We know that 2tan⁻¹x = tan⁻¹(2x/(1-x²))Here, x = 1/5= tan⁻¹(2 × (1/5) / (1 - (1/5)²))= tan⁻¹(2/5 / (1 - 1/25))= tan⁻¹(2/5 / (24/25))= tan⁻¹(2/5 × 25/24)= tan⁻¹(5/12)= 5/12