Find-the-positive-integer-n-such-that-tan-1-1-3-tan-1-1-4-tan-1-1-5-tan-1-1-n-pi-4-

Question Number 111725 by Aina Samuel Temidayo last updated on 04/Sep/20

Find the positive integer n such that

\tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{4}) + \tan^{- 1} (\frac{1}{5}) + \tan^{- 1} (\frac{1}{n}) = \frac{π}{4}

Answered by $@y@m last updated on 04/Sep/20

L e t \tan^{- 1} (\frac{1}{3}) = α \Rightarrow \tan α = \frac{1}{3}

L e t \tan^{- 1} (\frac{1}{4}) = β \Rightarrow \tan β = \frac{1}{4}

L e t \tan^{- 1} (\frac{1}{5}) = γ \Rightarrow \tan γ = \frac{1}{5}

L e t \tan^{- 1} (\frac{1}{n}) = δ \Rightarrow \tan δ = \frac{1}{n}

A T Q,

α + β + γ + δ = \frac{π}{4}

α + β + γ = \frac{π}{4} - δ

\tan (α + β + γ) = \tan (\frac{π}{4} - δ)

\frac{\frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{3.4.5}}{1 - \frac{1}{3.4} - \frac{1}{4.5} - \frac{1}{5.3}} = \frac{1 - \tan δ}{1 + \tan δ}

\frac{1 - \tan δ}{1 + \tan δ} = \frac{\frac{20 + 15 + 12 - 1}{60}}{\frac{60 - 5 - 3 - 4}{60}} = \frac{46}{48} = \frac{23}{24}

24 (1 - \tan δ) = 23 (1 + \tan δ)

1 = 47 \tan δ

\tan δ = \frac{1}{47}

δ = \tan^{- 1} (\frac{1}{47})

n = 47

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

47 is part of the options I have here .

Thanks .