Question

the value of (tan1. tan2. tan3.......... tan89) is​

Answer 1

Step-by-step explanation:

Note that

tan(90∘−A)=cotAhence tanAtan(90∘−A)=tanAcotA=tanA×1tanA=1

Now we can writetan1°tan2°tan3°...tan44°tan45°tan46°...tan87°tan88°tan89as=(tan1∘tan89∘)×(tan2∘tan88∘)×(tan3∘tan87∘)....(tan44∘tan46∘)×tan45∘=(tan1∘tan(90−1)∘)×(tan2∘tan(90−2)∘)×(tan3∘tan(90−3)∘)...(tan44∘tan(90−44)∘)×tan45∘= 1×1×1×.......×1×1= 1