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Solve the Triangle

Source: 1993 National High School Mathematics League, Exam One, Problem 5

February 27, 2020

Problem Statement

In △ABC\triangle ABC, c−ac-a is equal to height on side ACAC. Then, the value of sin⁡C−A2+cos⁡C+A2\sin\frac{C-A}{2}+\cos\frac{C+A}{2} is (A)1(B)12(C)13(D)−1\text{(A)}1\qquad\text{(B)}\frac{1}{2}\qquad\text{(C)}\frac{1}{3}\qquad\text{(D)}-1
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