The value of the expression $\dfrac{\sqrt{5}-2}{\sqrt{5}+2}-\dfrac{\sqrt{5}+2}{\sqrt{5}-2}$ is
[a] 8
[b] 25
[c] $-8\sqrt{5}$
[d] $8\sqrt{5}$

Hint: Rationalise the denominator of $\dfrac{\sqrt{5}+2}{\sqrt{5}-2}$ by multiplying numerator and denominator by $\sqrt{5}+2$ and use the identities ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ and $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to simplify the expression. Rationalise the denominator of $\dfrac{\sqrt{5}-2}{\sqrt{5}+2}$ by multiplying numerator and denominator by $\sqrt{5}-2$ and use the identities ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ to simplify the expression. Hence find the value of $\dfrac{\sqrt{5}-2}{\sqrt{5}+2}-\dfrac{\sqrt{5}+2}{\sqrt{5}-2}$.

Complete step-by-step answer:
Now, let us start the solution by simplifying the first term, $\dfrac{\sqrt{5}+2}{\sqrt{5}-2}$
Multiplying numerator and denominator by $\sqrt{5}+2$, we get
$\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\times \dfrac{\sqrt{5}+2}{\sqrt{5}+2}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Putting $a=\sqrt{5}$ and $b=2$, we get
$\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)={{\left( \sqrt{5} \right)}^{2}}-{{2}^{2}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$.
Hence, we have
$\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)=5-4=1$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Putting $a=\sqrt{5}$ and $b=2$, we get
${{\left( \sqrt{5}+2 \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}+2\times \sqrt{5}\times 2+{{2}^{2}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$.
Hence, we have
${{\left( \sqrt{5}+2 \right)}^{2}}=5+4\sqrt{5}+4=9+4\sqrt{5}$
Hence, we have
$\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=\dfrac{9+4\sqrt{5}}{1}=9+4\sqrt{5}$
Now, let us consider the second term $\dfrac{\sqrt{5}-2}{\sqrt{5}+2}$ and simplify it.
Multiplying numerator and denominator by $\sqrt{5}-2$, we get
$\dfrac{\sqrt{5}-2}{\sqrt{5}+2}=\dfrac{\sqrt{5}-2}{\sqrt{5}+2}\times \dfrac{\sqrt{5}-2}{\sqrt{5}-2}$
We know that $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
Put $a=\sqrt{5}$ and $b=2$, we get
$\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)={{\left( \sqrt{5} \right)}^{2}}-{{2}^{2}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$.
Hence, we have
$\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)=5-4=1$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Put $a=\sqrt{5}$ and $b=2$, we get
${{\left( \sqrt{5}-2 \right)}^{2}}={{\left( \sqrt{5} \right)}^{2}}-2\times \sqrt{5}\times 2+{{2}^{2}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$.
Hence, we have
${{\left( \sqrt{5}+2 \right)}^{2}}=5-4\sqrt{5}+4=9-4\sqrt{5}$
Hence, we have
$\dfrac{\sqrt{5}-2}{\sqrt{5}+2}=\dfrac{9-4\sqrt{5}}{1}=9-4\sqrt{5}$
Hence, we have
$\dfrac{\sqrt{5}-2}{\sqrt{5}+2}-\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=\left( 9-4\sqrt{5} \right)-\left( 9+4\sqrt{5} \right)=-8\sqrt{5}$
Hence option [c] is correct.

Note: [1] Alternative solution:
We know that $\dfrac{a}{b}-\dfrac{b}{a}=\dfrac{{{a}^{2}}-{{b}^{2}}}{ab}=\dfrac{\left( a+b \right)\left( a-b \right)}{ab}$
Put $a=\sqrt{5}-2$ and $b=\sqrt{5}+2$, we get
$\begin{align}
  & \dfrac{\sqrt{5}-2}{\sqrt{5}+2}-\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=\dfrac{\left( \sqrt{5}-2-\left( \sqrt{5}+2 \right) \right)\left( \sqrt{5}-2+\sqrt{5}+2 \right)}{\left( \sqrt{5}+2 \right)\left( \sqrt{5}-2 \right)} \\
 & =\dfrac{-4\left( 2\sqrt{5} \right)}{1}=-8\sqrt{5} \\
\end{align}$
Which is the same as obtained above.
Hence option [c] is correct.
Alternative Solution:
We know that $\dfrac{a-\sqrt{b}}{a+\sqrt{b}}-\dfrac{a+\sqrt{b}}{a-\sqrt{b}}=\dfrac{-4a\sqrt{b}}{{{a}^{2}}-b}$
Put $a=2$ and $b=5$, we get
$\dfrac{2-\sqrt{5}}{2+\sqrt{5}}-\dfrac{2+\sqrt{5}}{2-\sqrt{5}}=\dfrac{-4\left( 2 \right)\left( \sqrt{5} \right)}{4-5}$
Multiplying both sides by -1, we get
$\dfrac{\sqrt{5}-2}{\sqrt{5}+2}-\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=-8\sqrt{5}$, which is the same as obtained above.
Hence option [c] is correct.