Answer :
[tex]\sin(x) = \cos(90-x) <=> \frac{sin(x)}{cos(90-x)} = 1\\\prod_{n=1}^{89}\tan(n) = \frac{\sin(1)}{\cos(1)} \cdot \frac{sin(2)}{cos(2)}...\frac{sin(89)}{cos(89)} = 1[/tex]
[tex]\sin(x) = \cos(90-x) <=> \frac{sin(x)}{cos(90-x)} = 1\\\prod_{n=1}^{89}\tan(n) = \frac{\sin(1)}{\cos(1)} \cdot \frac{sin(2)}{cos(2)}...\frac{sin(89)}{cos(89)} = 1[/tex]