2{tan ^{ - 1}}left( {frac{1}{3}} right) + {ta | vedclass

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(D) We use the formula $2{	an ^{ - 1}}(x) = {	an ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} ight)$ for $|x| < 1$. First,simplify $2{	an ^{ - 1}}\l

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$\frac{\pi }{4}$

Vedclass · Answer

(D) We use the formula $2{	an ^{ - 1}}(x) = {	an ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} ight)$ for $|x| First,simplify $2{	an ^{ - 1}}\left( {\frac{1}{3}} ight)$: $2{	an ^{ - 1}}\left( {\frac{1}{3}} ight) = {	an ^{ - 1}}\left( {\frac{{2(1/3)}}{{1 - {{(1/3)}^2}}}} ight) = {	an ^{ - 1}}\left( {\frac{{2/3}}{{1 - 1/9}}} ight) = {	an ^{ - 1}}\left( {\frac{{2/3}}{{8/9}}} ight) = {	an ^{ - 1}}\left( {\frac{2}{3} 	imes \frac{9}{8}} ight) = {	an ^{ - 1}}\left( {\frac{3}{4}} ight)$. Now,add the remaining term using the formula ${	an ^{ - 1}}(x) + {	an ^{ - 1}}(y) = {	an ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} ight)$: ${	an ^{ - 1}}\left( {\frac{3}{4}} ight) + {	an ^{ - 1}}\left( {\frac{1}{7}} ight) = {	an ^{ - 1}}\left( {\frac{{3/4 + 1/7}}{{1 - (3/4)(1/7)}}} ight)$. Calculate the numerator and denominator: Numerator: $\frac{3}{4} + \frac{1}{7} = \frac{{21 + 4}}{{28}} = \frac{{25}}{{28}}$. Denominator: $1 - \frac{3}{{28}} = \frac{{28 - 3}}{{28}} = \frac{{25}}{{28}}$. Thus,${	an ^{ - 1}}\left( {\frac{{25/28}}{{25/28}}} ight) = {	an ^{ - 1}}(1) = \frac{\pi }{4}$.

$2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) = $

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