Question

${ \tan }^{ - 1} ( \frac{x + 1}{x + 3} ) + {tan}^{ - 1} ( \frac{x - 2}{x + 4} ) = \frac{\pi}{4}$
Solve it :​

Answer 1

${\fbox{\boxed {\huge{\rm{\red{Answer}}}}}}$

Step-by-step explanation:

$\rightarrow \: \large\mathtt{ { tan }^{ - 1} } \: \Large\mathtt{[ \frac { \frac {x + 1}{x + 3} + \frac{x - 2}{x + 4} }{1 - (\frac{x + 1 }{x + 3}) (\frac{x - 2}{ x + 4}) } ]} = \Large \mathtt{ \frac{\pi}{4} }$

$\rightarrow \: \large \mathtt{ {tan}^{ - 1} } \: \Large\mathtt{ [ \frac{ \frac{(x + 1)(x + 4) + (x - 2)(x + 3)}{(x + 3)(x + 4)} }{ \frac{(x + 3)(x + 4) - (x + 1)(x - 2)}{(x + 3)(x + 4)} } ] } = \large \mathtt{\frac{\pi}{4}}$

$\rightarrow \: \large \mathtt{ {tan}^{ - 1}} \: \large \mathtt{ [ \frac{ {(x}^{2} +4x + x + 4) + ( {x}^{2} + 3x - 2x - 6 }{( {x}^{2} + 4x + 3x + 12) - ( {x}^{2} - 2x + x - 2) } ] } = \large \mathtt{ \frac{\pi}{4}}$

$\rightarrow \: \large \mathtt{ {tan}^{ - 1} } \: \large \mathtt{ [ \frac{ {2x}^{2} + 6x - 2 }{8x + 14} ] } = \large \mathtt{ \frac{\pi}{4}}$

$\rightarrow \: \large \mathtt{tan} \: \large \mathtt{ [ {tan}^{ - 1} \: \frac{ {2x}^{2} + 6x - 2 }{8x + 14} ] } = \large \mathtt{tan \frac{\pi}{4}}$

$\rightarrow \: \large \mathtt{ \frac{ {2x}^{2} + 6x - 2 }{8x + 14} = 1}$

$\rightarrow \: \large \mathtt{ {2x}^{2 } + 6x - 2 = 8x + 14}$

$\rightarrow \: \large \mathtt{ {2x}^{2} - 2x - 16 = 0}$

$\rightarrow \: \large \mathtt{D = {b}^{2} - 4ac } \\ \: \: \: \: \: \: \: \: \: \: \: = \mathtt{(1) - 4(1)( - 8)} \\ \: \: \: \: \: \: \: \: \: \: \: = \mathtt{ \sqrt{33}}$

$\rightarrow \: \large\fbox\mathtt{ \alpha = \frac{1 \pm \sqrt{33} }{2} }$

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