Question

Solve $\tan 1^{\circ }\tan 2^{\circ }...\tan {89}^{\circ }=$

Answer 1

We have, $\tan 1^{\circ }\tan 2^{\circ }...\tan {89}^{\circ }$

$=\left(\tan 1^{\circ }\tan {89}^{\circ }\right)\left(\tan 2^{\circ }\tan {88}^{\circ }\right)...\left(\tan {44}^{\circ }\tan {46}^{\circ }\right)\tan {45}^{\circ }$

$=\left[\tan 1^{\circ }\tan ({90}^{\circ }-1^{\circ })\right]\left[\tan 2^{\circ }\tan ({90}^{\circ }-2^{\circ })\right]...\left[\tan {44}^{\circ }\tan ({90}^{\circ }-{44}^{\circ })\right]\tan {45}^{\circ }$

$=\left(\tan 1^{\circ }\cot 1^{\circ }\right)\left(\tan 2^{\circ }\cot 2^{\circ }\right)...\left(\tan {44}^{\circ }\cot {44}^{\circ }\right)\tan {45}^{\circ }$

$=1$ [ $\because \tan \theta \cot \theta =1$ and $\tan {45}^{\circ }=1$]