Question

Plzz answer the question 5 fast

Answer 1

◢GIVEN

$= > f(x) = log[ \frac{1 - x}{1 + x} ]$
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◢THEN,

$= > f(a) = log( \frac{ 1 - a}{1 + a} )$

◢AND

$= > f(b) = log( \frac{1 - b}{1 + b} )$
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◢NOW

$= > f(a) + f(b) = log[ \frac{1 - a}{1 + a} ] + log[ \frac{1 - b}{1 + b} ]$
______________________[◢Eq(1)]

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● WE CAN USE HERE :-

$= > log \alpha + log \beta = log( \alpha . \beta )$
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◢BY Eq(1)

$= > f(a) + f(b) = log[ \frac{(1 - a)(1 - b)}{(1 + a)(1 + b)} ]$
________________________[◢Eq(2)]

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◢NOW

$= > f(x) = log[ \frac{1 - x}{1 + x} ] \\ \\ = > f[ \frac{a + b}{1 + ab} ] = log [\frac{1 - \frac{a + b}{1 + ab} }{1 + \frac{a + b}{1 + ab} } ] \\ \\ = > f( \frac{a + b}{1 + ab} ) = log[ \frac{1 + ab - a - b}{1 + ab + a + b} ] \\ \\ = > f( \frac{a + b}{1 + ab} ) = log[ \frac{a(b - 1) - 1(b - 1)}{a(b + 1) + 1(b + 1)} ] \\ \\ = > f( \frac{a + b}{1 + ab} ) = log [\frac{(a - 1)(b - 1)}{(a + 1)(b + 1)} ] \\ \\ = > f[ \frac{a + b}{1 + ab} ] = log[ \frac{(1 - a)(1 - b)}{(1 + a)(1 + b)} ]$
_________________________[◢Eq(3)]

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◢NOW

● BY Eq(2) AND Eq(3)

$= > f(a) + f(b) = f[ \frac{a + b}{1 + ab} ]$
_____________________[◢PROVED]

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