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Show that:$ \frac{\left(a+\frac{1}{b}\right)^{m} \times\left(a-\frac{1}{b}\right)^{n}}{\left(b+\frac{1}{a}\right)^{m} \times\left(b-\frac{1}{a}\right)^{n}}=\left(\frac{a}{b}\right)^{m+n} $
Given:
\( \frac{\left(a+\frac{1}{b}\right)^{m} \times\left(a-\frac{1}{b}\right)^{n}}{\left(b+\frac{1}{a}\right)^{m} \times\left(b-\frac{1}{a}\right)^{n}}=\left(\frac{a}{b}\right)^{m+n} \)
To do:
We have to show that \( \frac{\left(a+\frac{1}{b}\right)^{m} \times\left(a-\frac{1}{b}\right)^{n}}{\left(b+\frac{1}{a}\right)^{m} \times\left(b-\frac{1}{a}\right)^{n}}=\left(\frac{a}{b}\right)^{m+n} \).
Solution:
We know that,
$(a^{m})^{n}=a^{m n}$
$a^{m} \times a^{n}=a^{m+n}$
$a^{m} \div a^{n}=a^{m-n}$
$a^{0}=1$
Therefore,
LHS $=\frac{(a+\frac{1}{b})^{m} \times (a-\frac{1}{b})^{n}}{(b+\frac{1}{a})^{m} \times (b-\frac{1}{a})^{n}}$
$=\frac{(\frac{a b+1}{b})^{m} \times (\frac{a b-1}{b})^{n}}{(\frac{a b+1}{a})^{m} \times (\frac{a b-1}{a})^{n}}$
$=\frac{\frac{(a b+1)^{m} \times(a b-1)^{n}}{b^{m} \times b^{n}}}{\frac{(a b+1)^{m}}{a^{m}} \times \frac{(a b-1)^{n}}{a^{n}}}$
$=\frac{(a b+1)^{m}(a b-1)^{n} \times a^{m} \times a^{n}}{b^{m} \times b^{n}(a b+1)^{m}(a b-1)^{n}}$
$=\frac{a^{m} \times a^{n}}{b^{m} \times b^{n}}$
$=\frac{a^{m+n}}{b^{m+n}}$
$=(\frac{a}{b})^{m+n}$
$=$ RHS
Hence proved.
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