Question
Prove that:
$\frac{1}{1+\text{x}^{\text{a}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}}=1$
$\frac{1}{1+\text{x}^{\text{a}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}}=1$
✓
Answer
$\frac{1}{1+\text{x}^{\text{a}-\text{b}}}+\frac{1}{1+\text{x}^{\text{b}-\text{a}}}=1$
Left hand side (LHS) = Right hand side (RHS) Considering LHS,
$=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}}$
$=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}}+\frac{\text{x}^\text{a}}{\text{x}^2+\text{x}^\text{b}}$
$=\frac{\text{x}^\text{b}+\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}\text{b}}$
$=1$
Therefore, LHS = RHS Hence proved
Left hand side (LHS) = Right hand side (RHS) Considering LHS,
$=\frac{1}{1+\frac{\text{x}^\text{a}}{\text{x}^\text{b}}}+\frac{1}{1+\frac{\text{x}^\text{b}}{\text{x}^\text{a}}}$
$=\frac{\text{x}^\text{b}}{\text{x}^\text{b}+\text{x}^\text{a}}+\frac{\text{x}^\text{a}}{\text{x}^2+\text{x}^\text{b}}$
$=\frac{\text{x}^\text{b}+\text{x}^\text{a}}{\text{x}^\text{a}+\text{x}\text{b}}$
$=1$
Therefore, LHS = RHS Hence proved
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