The numerical value of tan (2 tan^(−1) (1/5) − (π/4)) is


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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

	${t a n}^{- 1} (\frac{1}{5}) + {t a n}^{- 1} (\frac{1}{5}) = {t a n}^{- 1} (\frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \times \frac{1}{5}}) = {t a n}^{- 1} (\frac{\frac{2}{5}}{1 - \frac{1}{25}})$ $= {t a n}^{- 1} (\frac{2}{5} \times \frac{25}{24}) = {t a n}^{- 1} (\frac{5}{12})$ $t a n ({t a n}^{- 1} (\frac{5}{12}) - \frac{π}{4}) = \frac{t a n ({t a n}^{- 1} (\frac{5}{12})) - t a n (\frac{π}{4})}{1 + t a n ({t a n}^{- 1} (\frac{5}{12})) t a n (\frac{π}{4})}$ $= \frac{\frac{5}{12} - 1}{1 + \frac{5}{12} \times 1} = \frac{- 7}{17} p l s c h e c k . . .$
	Answered by hknkrc46 last updated on 02/Dec/18
	$tan (2tan^(−1) (1/5)−(𝛑/4))=((tan (2tan^(−1) (1/5))−tan (𝛑/4))/(1+tan (2tan^(−1) (1/5))tan (𝛑/4))) =((tan (tan^(−1) (1/5)+tan^(−1) (1/5))−tan (𝛑/4))/(1+tan (tan^(−1) (1/5)+tan^(−1) (1/5))tan (𝛑/4))) =((((tan (tan^(−1) (1/5))+tan (tan^(−1) (1/5)))/(1−tan (tan^(−1) (1/5))tan (tan^(−1) (1/5))))−tan (𝛑/4))/(1+((tan (tan^(−1) (1/5))+tan (tan^(−1) (1/5)))/(1−tan (tan^(−1) (1/5))tan (tan^(−1) (1/5))))tan (𝛑/4))) =(((((1/5)+(1/5))/(1−(1/5)∙(1/5)))−1)/(1+(((1/5)+(1/5))/(1−(1/5)∙(1/5)))))=((((2/5)/((24)/(25)))−1)/(1+((2/5)/((24)/(25)))))=(((5/(12))−1)/(1+(5/(12))))=((−(7/(12)))/((17)/(12)))=−(7/(17)) ★ {tan (tan^(−1) (1/5))=(1/5) ; tan (𝛑/4)=1}$
	$\tan ({2 \tan}^{- 1} \frac{1}{5} - \frac{π}{4}) = \frac{\tan ({2 \tan}^{- 1} \frac{1}{5}) - \tan \frac{π}{4}}{1 + \tan ({2 \tan}^{- 1} \frac{1}{5}) \tan \frac{π}{4}}$ $= \frac{\tan (\tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{5}) - \tan \frac{π}{4}}{1 + \tan (\tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{5}) \tan \frac{π}{4}}$ $= \frac{\frac{\tan (\tan^{- 1} \frac{1}{5}) + \tan (\tan^{- 1} \frac{1}{5})}{1 - \tan (\tan^{- 1} \frac{1}{5}) \tan (\tan^{- 1} \frac{1}{5})} - \tan \frac{π}{4}}{1 + \frac{\tan (\tan^{- 1} \frac{1}{5}) + \tan (\tan^{- 1} \frac{1}{5})}{1 - \tan (\tan^{- 1} \frac{1}{5}) \tan (\tan^{- 1} \frac{1}{5})} \tan \frac{π}{4}}$ $= \frac{\frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \cdot \frac{1}{5}} - 1}{1 + \frac{\frac{1}{5} + \frac{1}{5}}{1 - \frac{1}{5} \cdot \frac{1}{5}}} = \frac{\frac{\frac{2}{5}}{\frac{24}{25}} - 1}{1 + \frac{\frac{2}{5}}{\frac{24}{25}}} = \frac{\frac{5}{12} - 1}{1 + \frac{5}{12}} = \frac{- \frac{7}{12}}{\frac{17}{12}} = - \frac{7}{17}$ $★ {\tan (\tan^{- 1} \frac{1}{5}) = \frac{1}{5}; \tan \frac{π}{4} = 1}$
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