Find the value of \[a - \dfrac{1}{a}\], if \[a = 2 + \sqrt 3\]

Hint: First find the value of \[\dfrac{1}{a}\] by rationalisation then find the value then put it in \[a - \dfrac{1}{a}\] which will ultimately give you the answer.

Complete step-by-step answer:
Let us try to find the value of \[\dfrac{1}{a}\] we know that \[a = 2 + \sqrt 3 ,\] therefore we will put it to get the value of \[\dfrac{1}{a}\]
\[\therefore \dfrac{1}{a} = \dfrac{1}{{2 + \sqrt 3 }}\]
Now if we do rationalisation here i.e., if we multiply both the numerator and denominator with \[2 - \sqrt 3 ,\] we will get the rationalised form
\[\begin{array}{l}
 = \dfrac{1}{{2 + \sqrt 3 }} \times \dfrac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }}\\
 = \dfrac{{2 - \sqrt 3 }}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}\\
 = \dfrac{{2 - \sqrt 3 }}{{{2^2} - {{\left( {\sqrt 3 } \right)}^2}}}\\
 = \dfrac{{2 - \sqrt 3 }}{{4 - 3}}\\
 = 2 - \sqrt 3
\end{array}\]
So now we have the value of \[\dfrac{1}{a} = 2 - \sqrt 3 \]
Let us put it in \[a - \dfrac{1}{a}\] and find out the required value
\[\begin{array}{l}
\therefore a - \dfrac{1}{a}\\
 = 2 + \sqrt 3 + 2 - \sqrt 3 \\
 = 2 + 2\\
 = 4
\end{array}\]
So we have the value of \[a - \dfrac{1}{a}\] as 4.

Note: While rationalising we have used an algebraic formula \[{a^2} - {b^2} = (a - b)(a + b)\]
The denominator also notes that if we put the value of a directly in \[a - \dfrac{1}{a}\] and solve it by taking LCM the denominator will contain the root at last and we need to rationalise it there also.