Question

can any one solve this question please ​

Answer 1

Answer:

(α+β)=π/4

Step-by-step explanation:

We have,

$\tan( \alpha ) = \frac{1}{1 + {2}^{ - x} } \: \: and \: \: \tan( \beta ) = \frac{1}{1 + {2}^{x + 1} }$

Now, we have,

$\tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }$

$= > \tan( \alpha + \beta ) = \frac{ \frac{1}{1 + {2}^{ - x} } + \frac{1}{1 + {2}^{x + 1} } }{1 - \frac{1}{(1 + {2}^{ - x})(1 + {2}^{x + 1}) } }$

$= > \tan( \alpha + \beta ) = \frac{ \frac{ {2}^{x} }{1 + {2}^{x} } + \frac{1}{1 + {2}^{x + 1} } }{1 - \frac{ {2}^{x} }{(1 + {2}^{x})(1 + {2}^{x + 1}) } }$

$= > \tan( \alpha + \beta ) = \frac{ \frac{ {2}^{x} (1 + {2}^{x + 1}) + (1 + {2}^{x}) }{(1 + {2}^{x} )(1 + {2}^{x + 1} )} }{ \frac{(1 + {2}^{x})(1 + {2}^{x + 1}) - {2}^{x} }{(1 + {2}^{x})(1 + {2}^{x + 1} ) } }$

$= > \tan( \alpha + \beta ) = \frac{{2}^{x} + {2}^{2x + 1} + {2}^{x} + 1 }{1 + {2}^{x + 1} + {2}^{x} + {2}^{2x + 1} - {2}^{x} }$

$= > \tan( \alpha + \beta ) = \frac{ {2}^{x + 1} + {2}^{2x + 1} + 1}{ {2}^{x + 1} + {2}^{2x + 1} + 1}$

$= > \tan( \alpha + \beta ) = 1 = > \alpha + \beta = \frac{\pi}{4}$