Question

Solve this one for me...

Evaluate
tan^(-1) (4)/(5) + tan^(-1) (5)/(7) - tan^(-1) (5)/(6) ​

Answer 1

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

$\rm :\longmapsto\: {tan}^{ - 1}\dfrac{4}{5} + {tan}^{ - 1}\dfrac{5}{7} - {tan}^{ - 1} \dfrac{5}{6}$

We know,

$\boxed{ \sf{ \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x + y}{1 - xy} \bigg)}}$

So, bybusing this identity,

$\rm \: = \: \: \: {tan}^{ - 1}\bigg(\dfrac{\dfrac{4}{5} + \dfrac{5}{7} }{1 - \dfrac{4}{5} \times \dfrac{5}{7} } \bigg) - {tan}^{ - 1} \dfrac{5}{6}$

$\rm \: = \: \: \: {tan}^{ - 1}\bigg(\dfrac{\dfrac{28 + 25}{35} }{ \dfrac{35 - 20}{35} } \bigg) - {tan}^{ - 1} \dfrac{5}{6}$

$\rm \: = \: \: \: {tan}^{ - 1}\bigg(\dfrac{53 }{ 15} \bigg) - {tan}^{ - 1} \dfrac{5}{6}$

We know,

$\boxed{ \sf{ \: {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg(\dfrac{x - y}{1 + xy} \bigg)}}$

So, by applying this identity, we get

$\rm \: = \: \: \: {tan}^{ - 1}\bigg(\dfrac{\dfrac{53}{15} - \dfrac{5}{6} }{1 + \dfrac{53}{15} \times \dfrac{5}{6} } \bigg)$

$\rm \: = \: \: \: {tan}^{ - 1}\bigg(\dfrac{\dfrac{318 - 75}{90} }{ \dfrac{90 + 265}{90} } \bigg)$

$\rm \: = \: \: \: {tan}^{ - 1}\bigg(\dfrac{243}{355} \bigg)$

Additional information :-

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

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

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

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