Question
Evaluate : $\left(\frac{a}{2 b}+\frac{2 b}{a}\right)^2-\left(\frac{a}{2 b}-\frac{2 b}{a}\right)^2-4$
✓
Answer
Consider the given expression :
Let us expand the first term : $\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2$
We know that,
$(a+b)^2=a 2+b^2+2 a b$
$\therefore\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2=\left(\frac{a}{2 b}\right)^2+\left(\frac{2 b}{a}\right)^2+2 \times \frac{a}{2 b} \times \frac{2 b}{a}$
$=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}+2\ldots(1)$
Let us expand the second term : $\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2$
We know that,
$(a-b)^2=a^2+b^2-2 a b$
$ \therefore\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2=\left(\frac{a}{2 b}\right)^2+\left(\frac{2 b}{a}\right)^2-2 \times \frac{a}{2 b} \times \frac{2 b}{a}$
$=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}-2\ldots(2)$
Thus from $(1)$ and $(2)$, the given expression is
$\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2-\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2-4=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}+2-\frac{a^2}{(4 b)^2}-\frac{(4 b)^2}{a^2}+2-4$
$=0$.
Let us expand the first term : $\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2$
We know that,
$(a+b)^2=a 2+b^2+2 a b$
$\therefore\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2=\left(\frac{a}{2 b}\right)^2+\left(\frac{2 b}{a}\right)^2+2 \times \frac{a}{2 b} \times \frac{2 b}{a}$
$=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}+2\ldots(1)$
Let us expand the second term : $\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2$
We know that,
$(a-b)^2=a^2+b^2-2 a b$
$ \therefore\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2=\left(\frac{a}{2 b}\right)^2+\left(\frac{2 b}{a}\right)^2-2 \times \frac{a}{2 b} \times \frac{2 b}{a}$
$=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}-2\ldots(2)$
Thus from $(1)$ and $(2)$, the given expression is
$\left[\frac{a}{2 b}+\frac{2 b}{a}\right]^2-\left[\frac{a}{2 b}-\frac{2 b}{a}\right]^2-4=\frac{a^2}{(4 b)^2}+\frac{(4 b)^2}{a^2}+2-\frac{a^2}{(4 b)^2}-\frac{(4 b)^2}{a^2}+2-4$
$=0$.
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