Question

If $tanx^{-1} (x-1/x-2) + tanx^{-1} (x+1/x+2) = \pi /4$. Find the value of 'x'

Answer 1

Answer:

8 this is correct answer

Answer 2

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

Given equation is

$\rm :\longmapsto\: {tan}^{ - 1}\dfrac{x - 1}{x - 2} + {tan}^{ - 1}\dfrac{x + 1}{x + 2} = \dfrac{\pi}{4}$

We know that

$\boxed{ \bf{ \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1} \frac{x + y}{1 - xy}, \: \: provided \: that \: xy < 1}}$

So, using this result we get

$\rm :\longmapsto\:{tan}^{ - 1}\bigg(\dfrac{\dfrac{x - 1}{x - 2} + \dfrac{x + 1}{x + 2} }{1 - \dfrac{x - 1}{x - 2} \times \dfrac{x + 1}{x + 2} } \bigg) = \dfrac{\pi}{4}$

Provided that,

$\red{\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 }{ {x}^{2} - 4 } < 1}$

$\rm :\longmapsto\:\dfrac{\dfrac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)} }{\dfrac{(x - 2)(x + 2) - (x - 1)(x + 1)}{(x - 2)(x + 2)} } = tan\dfrac{\pi}{4}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 2x - x - 2 + {x}^{2} - x + 2x - 2}{ {x}^{2} - 4 - {x}^{2} + 1} = 1$

$\rm :\longmapsto\:\dfrac{ {2x}^{2} - 4 }{ - 3} = 1$

$\rm :\longmapsto\: {2x}^{2} - 4 = - 3$

$\rm :\longmapsto\: {2x}^{2} = - 3 + 4$

$\rm :\longmapsto\: {2x}^{2} = 1$

$\rm :\longmapsto\: {x}^{2} = \dfrac{1}{2}$

$\bf\implies \:x = \: \pm \: \dfrac{1}{ \sqrt{2} }$

Justification :-

$\rm :\longmapsto\:\dfrac{ {x}^{2} - 1 }{ {x}^{2} - 4 } < 1$

$\rm :\longmapsto\:\dfrac{\dfrac{1}{2} - 1}{\dfrac{1}{2} - 4} < 1$

$\rm :\longmapsto\:\dfrac{\dfrac{1 - 2}{2}}{\dfrac{1 - 8}{2}} < 1$

$\rm :\longmapsto\:\dfrac{ - 1}{ - 7} < 1$

$\rm :\longmapsto\:\dfrac{ 1}{ 7} < 1$

Hence, Justified

Additional Information :-

$\boxed{ \bf{ \: {tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1} \frac{x - y}{1 + xy}}}$

$\boxed{ \bf{ \: 2{tan}^{ - 1}x = {tan}^{ - 1} \frac{2x}{1 - {x}^{2} }}}$

$\boxed{ \bf{ \: 2{tan}^{ - 1}x = {sin}^{ - 1} \frac{2x}{1 + {x}^{2} }}}$

$\boxed{ \bf{ \: 2{tan}^{ - 1}x = {cos}^{ - 1} \frac{1 - {x}^{2} }{1 + {x}^{2} }}}$

$\boxed{ \bf{ \: 2 {cos}^{ - 1}x = {cos}^{ - 1}(2 {x}^{2} - 1)}}$

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

$\boxed{ \bf{ \: {2sin}^{ - 1}x = {cos}^{ - 1}(1 - {2x}^{2})}}$

$\boxed{ \bf{ \: {3cos}^{ - 1}x = {cos}^{ - 1}( {4x}^{3} - 3x)}}$