Question

Prove that $\tan 1^o\tan 2^o\tan 3^o....\tan {89}^o=1$

Answer 1

By complementary rule
$\left(\tan 1^o\cot 1^o\right)\left(\tan 2^o\cot 2^o\right)\left(\tan 3^o\cot 3^o\right).....\left(\tan {44}^o\cot {44}^o\right)\left(\tan {45}^o\right)=1$