Sean  y  las medidas de dos ángulos coterminales distintos, donde el lado final del ángulo  se encuentra en el tercer cuadrante. Si    ...

Para resolver este problema, primero debemos recordar que dos ángulos coterminales comparten el mismo lado inicial y final, pero difieren en múltiplos de 360 grados. Dado que el ángulo β se encuentra en el tercer cuadrante, sabemos que su tangente y cotangente serán negativas. Dado que α = 2a/3 * tan(4a/5) y β = 4a/1 * cot(2a), debemos calcular el valor de α5sen. Sustituyendo los valores dados, obtenemos: α = 2a/3 * tan(4a/5) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * (tan(4a)/tan(5)) = (2a/3) * 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