Question

Solve:
$\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\dfrac{2-\sqrt{3}}{2+\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

Answer 1

- Hint: Rationalise the denominator of each of the fractions. This can be done by multiplying numerator and denominator by the rational conjugate of the denominator. The rational conjugate of the number $a+\sqrt{b}$ is $a-\sqrt{b}$ and of the number $a-\sqrt{b}$ is $a+\sqrt{b}$. Finally, simplify the resulting expression and hence find the value of the given expression.

Complete step-by-step solution -

Rationalising the denominator of $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}$
Denominator $=2-\sqrt{3}$ which is of the form $a-\sqrt{b}$
Hence multiplying the numerator and denominator by $2+\sqrt{3}$, we get
$\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\left( \dfrac{2+\sqrt{3}}{2+\sqrt{3}} \right)$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above identity, we get
$\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{{{\left( 2+\sqrt{3} \right)}^{2}}}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$
Hence, we have
$\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=\dfrac{{{\left( 2+\sqrt{3} \right)}^{2}}}{4-3}={{\left( 2+\sqrt{3} \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using the above identity, we get
$\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=4+3+4\sqrt{3}=7+4\sqrt{3}$

Rationalising the denominator of $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$
Denominator $=\sqrt{3}+1$ which is of the form $a+\sqrt{b}$
Hence multiplying the numerator and denominator by $\sqrt{3}-1$, we get
$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\left( \dfrac{\sqrt{3}-1}{\sqrt{3}-1} \right)$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above identity, we get
$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$
Hence, we have
$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{3-1}=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{2}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using the above identity, we get
$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=\dfrac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}$
Rationalising the denominator of $\dfrac{2-\sqrt{3}}{2+\sqrt{3}}$
Denominator $=2+\sqrt{3}$ which is of the form $a+\sqrt{b}$
Hence multiplying the numerator and denominator by $2-\sqrt{3}$, we get
$\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\left( \dfrac{2-\sqrt{3}}{2-\sqrt{3}} \right)$
We know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
Using the above identity, we get
$\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\dfrac{{{\left( 2-\sqrt{3} \right)}^{2}}}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}$
We know that ${{\left( \sqrt{a} \right)}^{2}}=a$
Hence, we have
$\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\dfrac{{{\left( 2-\sqrt{3} \right)}^{2}}}{4-3}={{\left( 2-\sqrt{3} \right)}^{2}}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Using the above identity, we get
$\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=4+3-4\sqrt{3}=7-4\sqrt{3}$
Hence, the expression becomes
$7+4\sqrt{3}+7-4\sqrt{3}+2-\sqrt{3}=16-\sqrt{3}$
Hence the expression is equal to $16-\sqrt{3}$

Note: Alternatively, you can use \[\dfrac{a+\sqrt{b}}{a-\sqrt{b}}+\dfrac{a-\sqrt{b}}{a+\sqrt{b}}=\dfrac{2\left( {{a}^{2}}+b \right)}{{{a}^{2}}-b}\]
Put $a=2$ and $b=3$, we get
$\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\dfrac{2\left( 4+3 \right)}{4-3}=14$
Hence, the expression becomes
$14+2-\sqrt{3}=16-\sqrt{3}$, which is the same as obtained above.