Pls send it's ans and send with solution
Answers
Proof :-
Formula used :-
→ a² + b² + 2ab = (a+b)²
→ a² + b² - 2ab = (a - b)²
Explanation :-
Taking LHS
\begin{gathered} \to \sf f \bigg(\frac{2x}{1 + {x}^{2} } \bigg) \\ \\ hence \\ \\ \to \sf \log \bigg( \frac{1 + \frac{2x}{1 + {x}^{2} } }{1 - \frac{2x}{1 + {x}^{2} } } \bigg) \\ \\ \to \: \sf \log \bigg( \frac{ \frac{1 + {x}^{2} + 2x }{1 + {x}^{2} } }{ \frac{1 + {x}^{2} - 2x}{1 + {x}^{2} } } \bigg) \\ \\ \to \sf \: \log \bigg( \frac{ {x}^{2} + {1}^{2} + 2 \times 1 \times x }{ {x}^{2} + {1}^{2} - 2 \times 1 \times x} \bigg) \\ \\ \to \sf \log \bigg( \frac{ {(1 + x)}^{2} }{ {(1 - x)}^{2} } \bigg) \\ \\ \to \sf \: log{ \bigg( \frac{1 + x}{1 - x} \bigg) }^{2} \\ \\ \because \bf \: log( {m}^{n} ) = n \log m \\ \\ \to \sf \: 2 log \bigg( \frac{1 + x}{1 - x} \bigg) \\ \\ \to \sf \: 2f(x)\end{gathered}
→f(
1+x
2
2x
)
hence
→log(
1−
1+x
2
2x
1+
1+x
2
2x
)
→log(
1+x
2
1+x
2
−2x
1+x
2
1+x
2
+2x
)
→log(
x
2
+1
2
−2×1×x
x
2
+1
2
+2×1×x
)
→log(
(1−x)
2
(1+x)
2
)
→log(
1−x
1+x
)
2
∵log(m
n
)=nlogm
→2log(
1−x
1+x
)
→2f(x)