Question

What is the value of tan1o tan2o tan3o…………. tan 87o tan88o tan89o?

Answer 1

tan(1+2+3+........89)

Answer 2

@Auxuris

Answer 3

hint : \(\tan(\theta) = \cot(90 - \theta) = \dfrac{1}{\tan(90-\theta )}\)

Answer 4

\[\large \tan 1 \tan 2 \tan 3 \cdots \tan 87 \tan 88 \tan 89\]

Answer 5

focus on first and last terms

\[\large \color{Red}{\tan 1} \tan 2 \tan 3 \cdots \tan 87 \tan 88 \color{red}{\tan 89}\]

Answer 6

tan1=cot(90-1)=cot89

Answer 7

yes which equals ?

Answer 8

cot89.tan89=1

Answer 9

tan1=cot(90-1)=cot89 = 1/tan89

right ?

Answer 10

yep !

Answer 11

so basically we can pair up and cancel everything except the `middle term`
any idea what the middle term would be ?

Answer 12

hey can we do yesterdays problem

Answer 13

tan45=1

Answer 14

il msg u once i figure out how to solve yesterday's problem :)

Answer 15

using cos and sin terms instead of tan2x-sec2x=1