Question

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

$\large\purple\rightarrow$Mention the possible values of x, which satisfy the trigonometric equation $\rm{\tan ^{ - 1} ( \frac{x - 1}{x - 2}) + \tan ^{ - 1}( \frac{x + 1}{x + 2}) = \frac{\pi}{2}}$.​

Answer 1

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

Given that

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

We know that

$\boxed{ \bf{ \: {tan}^{ - 1}x + {tan}^{ - 1}y = {tan}^{ - 1}\bigg[ \frac{x + y}{1 - xy}\bigg] \: when \: xy \leqslant 1}}$

So, using this identity, we have

$\rm :\longmapsto\:{tan}^{ - 1}\bigg[ \dfrac{\dfrac{x - 1}{x - 2} + \dfrac{x + 1}{x + 2} }{1 - \dfrac{x - 1}{x - 2} \times \dfrac{x + 1}{x + 2} } \bigg] = \dfrac{\pi}{2}$

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

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

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

$\rm :\longmapsto\:{tan}^{ - 1}\bigg[\dfrac{ 2{x}^{2} - 4 }{ - 3}\bigg] = \dfrac{\pi}{2}$

$\rm :\longmapsto\:\bigg[\dfrac{ 2{x}^{2} - 4 }{ - 3}\bigg] = tan\dfrac{\pi}{2}$

$\rm :\longmapsto\:\bigg[\dfrac{ 2{x}^{2} - 4 }{ - 3}\bigg] = not \: defined$

$\bf\implies \: - 3 = 0$

$\rm :\longmapsto\:which \: is \: not \: possible$

$\bf\implies \:There \: is \: no \: real \: value \: of \: x$

Additional Information :-

$\rm :\longmapsto\:{tan}^{ - 1}x - {tan}^{ - 1}y = {tan}^{ - 1}\bigg[\dfrac{x - y}{1 + xy}\bigg]$

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

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

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

$\rm :\longmapsto\:2{sin}^{ - 1}x = {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) = {cos}^{ - 1}(1 - 2 {x}^{2})$

$\rm :\longmapsto\:2{cos}^{ - 1}x = {sin}^{ - 1}(2x \sqrt{1 - {x}^{2} }) = {cos}^{ - 1}(2 {x}^{2} - 1)$

Answer 2

*Question:—

Mention the possible values of x, which satisfy the trigonometric equation:—

→ tan^-1 (x-1/x-2) + tan^-1 (x+1)/(x+2) = π/2

*Answer:—

${ \tan}^{ - 1} ( \frac{x - 1}{x - 2} ) + { \tan}^{ - 1} ( \frac{x + 1}{x + 2} ) = \frac{\pi}{2}$

$= > { \tan}^{ - 1} [ \frac{ \frac{x - 1}{x - 2} + \frac{x + 1}{x + 2} }{1 - \frac{(x - 1)}{(x - 2)} \frac{(x + 1)}{(x + 2)} } ] = \frac{\pi}{2}$

$= > \frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{ {x}^{2} - 4 - ( {x}^{2} - 1)} = 1$

$= > \frac{ {x}^{2} + x - 2 + {x}^{2} - x - 2 }{ - 3} = 1$

$= > 2 {x}^{2} - 2 = - 3$

$= > {2x}^{2} = - 3 + 2$

$= > {2x}^{2} = - 1$

$= > x = \frac{ - 1}{ \sqrt{2} }$