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Exponents Of Real Numbers — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsExponents Of Real Numbers2 Marks
Question
Show that: $\frac{\Big(\text{a}+\frac{1}{\text{b}}\Big)^\text{m}\times\Big(\text{a}-\frac{1}{\text{b}}\Big)^\text{n}}{\Big(\text{b}+\frac{1}{\text{a}}\Big)^\text{m}\times\Big(\text{b}-\frac{1}{\text{a}}\Big)^\text{n}}=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}+\text{n}}$
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Answer

$\text{LHS}=\frac{\Big(\text{a}+\frac{1}{\text{b}}\Big)^\text{m}\times\Big(\text{a}-\frac{1}{\text{b}}\Big)^\text{n}}{\Big(\text{b}+\frac{1}{\text{a}}\Big)^\text{m}\times\Big(\text{b}-\frac{1}{\text{a}}\Big)^\text{n}}$
$=\frac{\Big(\frac{​\text{ab}+1​}{\text{b}}\Big)^\text{m}\Big(\frac{\text{ab}-1}{\text{b}}\Big)^\text{n}}{\Big(\frac{\text{ab}+1}{\text{a}}\Big)^\text{m}\Big(\frac{\text{ab}-1}{\text{a}}\Big)^{\text{n}}}$
$=\frac{\frac{\text{(ab}+1)^\text{m}}{\text{b}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{b}^\text{n}}}{\frac{(\text{ab}+1)^\text{m}}{\text{a}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{a}^\text{n}}}$
$=\frac{(\text{ab}+1)^\text{m}}{\text{b}^\text{m}}\times\frac{(\text{ab}-1)^\text{n}}{\text{b}^\text{n}}\times\frac{\text{a}^\text{m}}{(\text{ab}+1)^\text{m}}\times\frac{\text{a}^\text{n}}{(\text{ab}-1)^\text{n}}$
$=\frac{\text{a}^{\text{m}+\text{n}}}{\text{b}^{\text{m}+\text{n}}}$
$=\Big(\frac{\text{a}}{\text{b}}\Big)^{\text{m}+\text{n}}$
$=\text{RHS}$

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