Question

14. The value of (tan1° tan2° tan3º... tan89°) is
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Answer 1

Step-by-step explanation:

tan1°× tan2°× tan3°….tan89°

Pairing tan

= (tan1°× tan89°)×( tan2°tan88°)(tan3°× tan87°)×…….× (tan44° tan46°)tan45°

Using cot(90° - x) = tan x

= (tan1°× Cot(90-89°)×( tan2°Cot(90-88°)(tan3°× Cot(90-87°)×…….× (tan44° Cot(90-46°)tan45°

= (tan1°× cot 1°)×( tan2° cot 2°)(tan3°× cot 3°)×…….× (tan44° cot44°)tan45°

We know that,

tan x × cot x = tan x × 1/ tan x = 1 & tan45° =1

= 1 × 1 × 1 × …… × 1

= 1