Question

Question below

Answer 1

\[\log \tan1 + \log \tan2 + \log \tan3 + ........ + logtan89\]

Answer 2

I get a intuition we should group them

Answer 3

We can use this identity here :
log (a) + log(b) + ... + log(n)

log(ab...n)

Answer 4

So, it will become : log(tan(1) * tan(2) ... tan(89) )

Answer 5

0?

Answer 6

And then after tan (44) .. use this : \(\tan (\theta) = \cot (90 - \theta) \)
RIGHT!

Answer 7

I was gonna say all of that. but its ok

Answer 8

0 is correct OpenStudier. Good work

Answer 9

answer is zero the series can be written as
Log (tan(1)* tan(2) * .. * tan(89))
now tana=cot(90-a)
so the series is now\
Log( tan(1)* tan(2)*...* tan(44)* tan(45)*...* tan(88)* tan(89))
now replace tana by cota
Log( tan(1)* tan(2)*...* tan(44)* cot(44)*...* cot(2)* cot(1)) ===0

Answer 10

You were indeed correct

Answer 11

Thank you all of you !!!

Answer 12

Good Job openstudier

Answer 13

You're always welcome but half the job was done by you! :)

Answer 14

tan46 = tan(90-44) =cot 44
tan47 = tan(90-43) =cot 43
and so on
hence
tan1*tan2*.......................tan44*.tan45*tan46*tan47*.........................tan89
=tan1tan1*tan2*........................*tan44*tan45*cot44*.........................tan89
=tan1*cot1*tan2*cot2*....................................tan44*cot44*tan45
=1..
hence further...