prove-that-2tan-1-1-3-tan-1-1-7-pi-4-

Question Number 113200 by bobhans last updated on 11/Sep/20

prove that {2 \tan}^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{7}) = \frac{π}{4}

Answered by john santu last updated on 11/Sep/20

(Q) p r o v e t h a t 2 \tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{7}) = \frac{π}{4} .

(s o l) r e c a l l \tan^{- 1} (\frac{1}{3}) = a r g (3 + i)

\to 2 \tan^{- 1} (\frac{1}{3}) = 2 a r g (3 + i) = a r g ({(3 + i)}^{2}) = a r g (8 + 6 i)

= a r g (4 + 3 i)

n o w w e h a v e 2 \tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{7}) =

a r g (4 + 3 i) + a r g (7 + i) =

a r g ((4 + 3 i) (7 + i)) =

a r g (25 + 25 i) = a r g (1 + i) = \frac{π}{4} . (✓)

Answered by Dwaipayan Shikari last updated on 11/Sep/20

2 {t a n}^{- 1} \frac{1}{3} = {t a n}^{- 1} (\frac{\frac{1}{3} + \frac{1}{3}}{1 - \frac{1}{9}}) = {t a n}^{- 1} \frac{3}{4}

{t a n}^{- 1} (\frac{1}{7}) + {t a n}^{- 1} (\frac{3}{4}) = {t a n}^{- 1} (\frac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{3}{28}}) = {t a n}^{- 1} (1) = \frac{π}{4}