Question

If tan⁻¹ (x - 1) + tan⁻¹ (x) + tan⁻¹ ( x + 1 ) = tan⁻¹ (3x) then x =​

Answer 1

SOLUTION

TO SOLVE

$\displaystyle\sf{ { \tan}^{ - 1}(x - 1) + { \tan}^{ - 1}(x) + { \tan}^{ - 1}(x + 1) = { \tan}^{ - 1}(3x)}$

FORMULA TO BE IMPLEMENTED

$\displaystyle\sf{ { \tan}^{ - 1}(x ) + { \tan}^{ - 1}(y) = { \tan}^{ - 1} \bigg( \frac{x + y}{1 - xy} \bigg)}$

EVALUATION

$\displaystyle\sf{ { \tan}^{ - 1}(x - 1) + { \tan}^{ - 1}(x) + { \tan}^{ - 1}(x + 1) = { \tan}^{ - 1}(3x)}$

$\displaystyle\sf{ \implies { \tan}^{ - 1}(x - 1) + { \tan}^{ - 1}(x + 1) = { \tan}^{ - 1}(3x) - { \tan}^{ - 1}(x)}$

$\displaystyle\sf{ \implies { \tan}^{ - 1} \bigg( \frac{x - 1 + x + 1}{1 - ((x - 1)(x + 1))} \bigg) = { \tan}^{ - 1} \bigg( \frac{3x - x }{1 + 3x.x} \bigg) }$

$\displaystyle\sf{ \implies { \tan}^{ - 1} \bigg( \frac{2x}{2 - {x}^{2}} \bigg) = { \tan}^{ - 1} \bigg( \frac{2x }{1 + 3 {x}^{2} } \bigg) }$

$\displaystyle\sf{ \implies \bigg( \frac{2x}{2 - {x}^{2}} \bigg) = \bigg( \frac{2x }{1 + 3 {x}^{2} } \bigg) }$

$\displaystyle\sf{ \implies \: 2x \bigg( \frac{1}{2 - {x}^{2}} - \frac{1 }{1 + 3 {x}^{2} } \bigg) = 0}$

Now 2x = 0 gives x = 0

Again

$\displaystyle\sf{ \bigg( \frac{1}{2 - {x}^{2}} - \frac{1 }{1 + 3 {x}^{2} } \bigg) = 0 \: \: gives}$

$\displaystyle\sf{ \implies \: \frac{1}{2 - {x}^{2}} = \frac{1 }{1 + 3 {x}^{2} } \: }$

$\displaystyle\sf{ \implies \: 1 + 3 {x}^{2} = 2 - {x}^{2} }$

$\displaystyle\sf{ \implies \: 4 {x}^{2} = 1}$

$\displaystyle\sf{ \implies \: {x}^{2} = \frac{1}{4} }$

$\displaystyle\sf{ \implies \: x = \pm \: \frac{1}{2} }$

FINAL ANSWER

Hence the required solution is

$\boxed{ \: \: \displaystyle\sf{\: x= - \frac{1}{2} \: ,\: 0 \: ,\: \frac{1}{2} } \: \: }$

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