The-numerical-value-of-tan-2-tan-1-1-5-pi-4-is-

Question Number 49061 by Pk1167156@gmail.com last updated on 02/Dec/18

The numerical value of

\tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) is

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Dec/18

{t a n}^{- 1} (\frac{1}{5}) + {t a n}^{- 1} (\frac{1}{5}) = {t a n}^{- 1} (\frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \times \frac{1}{5}}) = {t a n}^{- 1} (\frac{\frac{2}{5}}{1 - \frac{1}{25}})

= {t a n}^{- 1} (\frac{2}{5} \times \frac{25}{24}) = {t a n}^{- 1} (\frac{5}{12})

t a n ({t a n}^{- 1} (\frac{5}{12}) - \frac{π}{4}) = \frac{t a n ({t a n}^{- 1} (\frac{5}{12})) - t a n (\frac{π}{4})}{1 + t a n ({t a n}^{- 1} (\frac{5}{12})) t a n (\frac{π}{4})}

= \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \times 1} = \frac{- 7}{17} p l s c h e c k \dots

Answered by hknkrc46 last updated on 02/Dec/18

\tan ({2 \tan}^{- 1} \frac{1}{5} - \frac{π}{4}) = \frac{\tan ({2 \tan}^{- 1} \frac{1}{5}) - \tan \frac{π}{4}}{1 + \tan ({2 \tan}^{- 1} \frac{1}{5}) \tan \frac{π}{4}}

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

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

= \frac{\frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \cdot \frac{1}{5}} - 1}{1 + \frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \cdot \frac{1}{5}}} = \frac{\frac{\frac{2}{5}}{\frac{24}{25}} - 1}{1 + \frac{\frac{2}{5}}{\frac{24}{25}}} = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12}} = \frac{- \frac{7}{12}}{\frac{17}{12}} = - \frac{7}{17}

★ {\tan (\tan^{- 1} \frac{1}{5}) = \frac{1}{5}; \tan \frac{π}{4} = 1}

Facebook Tweet Pin