Question

Show that: tan^-1(1/5) + tan^-1(1/7) + tan^-1(1/3) + tan^-1(1/8) = π * -24/13

Answer 1

Appropriate Question :-

Show that

$\rm :\longmapsto\: {tan}^{ - 1}\dfrac{1}{5} + {tan}^{ - 1}\dfrac{1}{7} + {tan}^{ - 1}\dfrac{1}{3} + {tan}^{ - 1}\dfrac{1}{8} = \dfrac{\pi}{4}$

$\large\underline{\sf{Solution-}}$

Consider,

$\rm :\longmapsto\: {tan}^{ - 1}\dfrac{1}{5} + {tan}^{ - 1}\dfrac{1}{7} + {tan}^{ - 1}\dfrac{1}{3} + {tan}^{ - 1}\dfrac{1}{8}$

$\rm \: = \: \bigg({tan}^{ - 1}\dfrac{1}{5} + {tan}^{ - 1}\dfrac{1}{7}\bigg) + \bigg({tan}^{ - 1}\dfrac{1}{3} + {tan}^{ - 1}\dfrac{1}{8}\bigg)$

We know,

$\boxed{ \sf{ \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\dfrac{x + y}{1 - xy} \: provided \: that \: xy < 1}}$

$\rm \:  =  \:  {tan}^{ - 1}\bigg(\dfrac{\dfrac{1}{5} + \dfrac{1}{7} }{1 - \dfrac{1}{5} \times \dfrac{1}{7} } \bigg) + {tan}^{ - 1}\bigg(\dfrac{\dfrac{1}{3} + \dfrac{1}{8} }{1 - \dfrac{1}{3} \times \dfrac{1}{8} } \bigg)$

$\rm \:  =  \:  \:{tan}^{ - 1}\dfrac{7 + 5}{35 - 1} + {tan}^{ - 1}\dfrac{8 + 3}{24 - 1}$

$\rm \:  =  \:  \:{tan}^{ - 1}\dfrac{12}{34} + {tan}^{ - 1}\dfrac{11}{23}$

$\rm \:  =  \:  \:{tan}^{ - 1}\dfrac{6}{17} + {tan}^{ - 1}\dfrac{11}{23}$

$\rm \:  =  \:  \:{tan}^{ - 1}\bigg(\dfrac{\dfrac{6}{17} + \dfrac{11}{23} }{1 - \dfrac{6}{17} \times \dfrac{11}{23} } \bigg)$

$\rm \:  =  \:  \:{tan}^{ - 1}\dfrac{138 + 187}{391 - 66}$

$\rm \:  =  \:  \:{tan}^{ - 1}\dfrac{325}{325}$

$\rm \:  =  \:  \:{tan}^{ - 1}1$

$\rm \:  =  \:  \:{tan}^{ - 1}\bigg(tan\dfrac{\pi}{4}\bigg)$

$\rm \:  =  \:  \:\dfrac{\pi}{4}$

Hence, Proved

Additional Information :-

$\boxed{ \sf{ \: {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}(x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} }}}$

$\boxed{ \sf{ \: {cos}^{ - 1}x - {cos}^{ - 1}y = {cos}^{ - 1}(xy+\sqrt{1 - {y}^{2}} \sqrt{1-{x}^{2}}}}$

$\boxed{ \sf{ \: {cos}^{ - 1}x + {cos}^{ - 1}y = {cos}^{ - 1}(xy - \sqrt{1 - {y}^{2}} \sqrt{1-{x}^{2}}}}$

$\boxed{ \sf{ \:2 {tan}^{ - 1}x = {sin}^{ - 1}\dfrac{2x}{1 + {x}^{2} }}}$

$\boxed{ \sf{ \:2 {tan}^{ - 1}x = {tan}^{ - 1}\dfrac{2x}{1 - {x}^{2} }}}$

$\boxed{ \sf{ \:2 {tan}^{ - 1}x = {cos}^{ - 1}\dfrac{1 - {x}^{2} }{1 + {x}^{2} }}}$