World number one Aryna Sabalenka has been chosen as the WTA player of the year, following a 2024 season in which she secured four titles, including two Grand Slams and two WTA 1000 events. The Belarusian athlete successfully defended her Australian Open title in January, subsequently clinching the US Open in September. This marked her third Grand Slam singles victory. The 26-year-old also achieved wins at the Cincinnati Open in August and the Wuhan Open in October. Her victory at the Wuhan Open made her the first woman to win that particular event three times. She ascended to the world number one ranking in October, concluding Iga Swiatek’s 11-month tenure at the top position. This was achieved despite her absence from Wimbledon due to injury. Additionally, Sabalenka chose not to participate in the Olympics in August, citing health prioritization. Her season concluded with a win-loss record of 56-14. This marks Sabalenka’s inaugural recognition as player of the year, a title previously claimed by Swiatek in both 2022 and 2023. International tennis media members cast votes to determine the winners. Post navigation Debate Arises Over James Tavernier’s Role and Future at Rangers Ireland Captain Doris Optimistic About Team Discipline and Debutants Ahead of Fiji Match