Question

$\tan ^{ - 1} ( \frac{x + 1}{x - 1} ) + \tan^{ - 1} ( \frac{x - 1}{x} ) = { \tan }^{ - 1} ( - 7)$
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Answer 1

GIVEN :–

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{x + 1}{x - 1} \right) + \tan^{ - 1} \left( \dfrac{x - 1}{x} \right) = { \tan }^{ - 1} ( - 7)$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

• We know that –

$\\ \implies \bf \tan ^{ - 1}(x) +\tan ^{ - 1}(y) = \tan ^{ - 1} \left( \dfrac{x+y}{1 -xy} \right)$

• So that –

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{\dfrac{x + 1}{x - 1} +\dfrac{x - 1}{x} }{1 -\dfrac{x + 1}{x - 1} \times \dfrac{x - 1}{x}} \right) = { \tan }^{ - 1} ( - 7)$

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{x(x + 1) + {(x - 1)}^{2}}{x(x - 1) - \{ {x}^{2} - 1\}}\right) = { \tan }^{ - 1} ( - 7)$

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{ {x}^{2} + x+ {x}^{2} + 1 -2x}{ {x}^{2} - x - {x}^{2} + 1}\right) = { \tan }^{ - 1} ( - 7)$

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{ 2{x}^{2}-x + 1 }{1 - x}\right) = { \tan }^{ - 1} ( - 7)$

• Now Let's compare –

$\\ \implies \bf \dfrac{ 2{x}^{2} -x+ 1}{1 - x}=- 7$

$\\ \implies \bf 2{x}^{2} -x+ 1=- 7(1 - x)$

$\\ \implies \bf 2{x}^{2}-x + 1=7x- 7$

$\\ \implies \bf 2{x}^{2} - 8x + 8 = 0$

$\\ \implies \bf {x}^{2} - 4x + 4 = 0$

$\\ \implies \bf {(x - 2)}^{2} = 0$

$\\ \implies \bf x = 2$

• Hence , The value of 'x' is 2.

Answer 2

Answer ⤵️

Given ⤵️

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{x + 1}{x - 1} \right) + \tan^{ - 1} \left( \dfrac{x - 1}{x} \right) = { \tan }^{ - 1} ( - 7)$

TO FIND ⤵️

☆Value of 'x' = ?

Solution ⤵️

☆We know:-

$\\ \implies \bf \tan ^{ - 1}(x) +\tan ^{ - 1}(y) = \tan ^{ - 1} \left( \dfrac{x+y}{1 -xy} \right)$

☆So that:-

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{\dfrac{x + 1}{x - 1} +\dfrac{x - 1}{x} }{1 -\dfrac{x + 1}{x - 1} \times \dfrac{x - 1}{x}} \right) = { \tan }^{ - 1} ( - 7)$

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{x(x + 1) + {(x - 1)}^{2}}{x(x - 1) - \{ {x}^{2} - 1\}}\right) = { \tan }^{ - 1} ( - 7)$

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{ {x}^{2} + x+ {x}^{2} + 1 -2x}{ {x}^{2} - x - {x}^{2} + 1}\right) = { \tan }^{ - 1} ( - 7)$

$\\ \implies \bf \tan ^{ - 1} \left( \dfrac{ 2{x}^{2}-x + 1 }{1 - x}\right) = { \tan }^{ - 1} ( - 7)$

☆now, comparing:-

$\\ \implies \bf \dfrac{ 2{x}^{2} -x+ 1}{1 - x}=- 7$

$\\ \implies \bf 2{x}^{2} -x+ 1=- 7(1 - x)$

$\\ \implies \bf 2{x}^{2}-x + 1=7x- 7$

$\\ \implies \bf 2{x}^{2} - 8x + 8 = 0$

$\\ \implies \bf {x}^{2} - 4x + 4 = 0$

$\\ \implies \bf {(x - 2)}^{2} = 0$

$\\ \implies \bf x = 2$

• Therefore ,The value of 'x' is 2.

☆Answer:- The value of 'x' is 2.

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