Question

prove that tan ^-1(1/7)+cot ^-1(13/1)=tan^1(2/9)​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\tt \: Prove \: that : \: {tan}^{ - 1} \dfrac{1}{7} + {cot}^{ - 1} 13 = {tan}^{ - 1} \dfrac{2}{9}$

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

Identities Used :-

$\boxed{ \bf \: {cot}^{ - 1} x = {tan}^{ - 1} \dfrac{1}{x} }$

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

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\: {tan}^{ - 1} \dfrac{1}{7} + {cot}^{ - 1}13$

$\: \: \sf \: = \: \: {tan}^{ - 1} \dfrac{1}{7} + {tan}^{ - 1} \dfrac{1}{13}$

$\: \: \sf \: = \: \: {tan}^{ - 1} \bigg(\dfrac{\dfrac{1}{7} + \dfrac{1}{13} }{ \: \: \: \: \: 1 - \dfrac{1}{7} \times \dfrac{1}{13} \: \: \: \: } \: \bigg)$

$\: \: \sf \: = \: \: {tan}^{ - 1} \bigg(\dfrac{\dfrac{13 + 7}{91} }{ \: \: \: \: \: 1 - \dfrac{1}{91} \: \: \: \: } \: \bigg)$

$\: \: \sf \: = \: \: {tan}^{ - 1} \bigg(\dfrac{\dfrac{20}{ \cancel{91}} }{ \: \: \: \: \:\dfrac{91 - 1}{ \cancel{91}} \: \: \: \: } \: \bigg)$

$\: \: \sf \: = \: \: {tan}^{ - 1} \dfrac{20}{90}$

$\: \: \sf \: = \: \: {tan}^{ - 1} \dfrac{2}{9}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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

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

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

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

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

$\boxed{ \sf \: {sec}^{ - 1} x + {cosec}^{ - 1} x = \dfrac{\pi}{2}}$