Question

Proof it by Rationalisation.

$\blue{ \frac{1}{ \sqrt{2} - \sqrt{3} - \sqrt{5} } + \frac{1}{ \sqrt{2} + \sqrt{3} - \sqrt{5} } = \pink{ \frac{1}{ \sqrt{2} } }}$

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Answer 1

Solution:-

To Prove:-

$\sf \dfrac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}=\dfrac{1}{\sqrt{2}}$

Taking LHS,

$\sf = \dfrac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}+\dfrac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$

Rationalising the denominator of the first fraction.

$\sf = \dfrac{1}{\left(\sqrt{2}-\sqrt{3}\right)-\sqrt{5}}$

$\sf = \dfrac{1}{\left(\sqrt{2}-\sqrt{3}\right)-\sqrt{5}}\times \dfrac{\left(\sqrt{2}-\sqrt{3}\right)+\sqrt{5}}{\left(\sqrt{2}-\sqrt{3}\right)+\sqrt{5}}$

$\sf = \dfrac{1\left(\left(\sqrt{2}-\sqrt{3}\right)+\sqrt{5}\right)}{\left(\left(\sqrt{2}-\sqrt{3}\right)-\sqrt{5}\right)\left(\left(\sqrt{2}-\sqrt{3}\right)+\sqrt{5}\right)}$

Use (a + b)(a - b) = a² - b² to simplify.

$\sf = \dfrac{\left(\sqrt{2}-\sqrt{3}\right)+\sqrt{5}}{\left(\sqrt{2}-\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}$

$\sf = \dfrac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\left(\sqrt{2}-\sqrt{3}\right)^{2}-5}$

Use (a - b)² = a² - 2ab + b² to expand the expression.

$\sf = \dfrac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2-2\sqrt{6}+3-5}$

$\sf = \dfrac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{-2\sqrt{6}}$

Rationalise it further.

$\sf = -\dfrac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2\sqrt{6}}$

$\sf = -\dfrac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\times \dfrac{\sqrt{6}}{\sqrt{6}}$

$\sf = -\dfrac{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\sqrt{6}}{2\sqrt{6}\sqrt{6}}$

$\sf = -\dfrac{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\sqrt{6}}{12}$

Rationalising the denominator of the second fraction.

$\sf = \dfrac{1}{\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}}$

$\sf = \dfrac{1}{\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}}\times \dfrac{\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}}$

$\sf = \dfrac{1\left(\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}\right)}{\left(\left(\sqrt{2}+\sqrt{3}\right)-\sqrt{5}\right)\left(\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}\right)}$

Use (a + b)(a - b) = a² - b² to simplify.

$\sf = \dfrac{\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}$

$\sf = \dfrac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}\right)^{2}-5}$

Use (a + b)² = a² + 2ab + b² to expand the expression.

$\sf = \dfrac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2+2\sqrt{6}+3-5}$

$\sf = \dfrac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}$

Rationalise it further.

$\sf = \dfrac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\times \dfrac{\sqrt{6}}{\sqrt{6}}$

$\sf = \dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\sqrt{6}}{2\sqrt{6}\sqrt{6}}$

$\sf = \dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\sqrt{6}}{12}$

Now, add both the fractions.

$\sf = -\dfrac{\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\sqrt{6}}{12}+\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\sqrt{6}}{12}$

Distribute $\sf \sqrt{6}$ through the parentheses.

$\sf = -\dfrac{\sqrt{12}-\sqrt{18}+\sqrt{30}}{12}+\dfrac{\sqrt{12}+\sqrt{18}+\sqrt{30}}{12}$

Write both the numerators above a common denominator.

$\sf = \dfrac{-\left(\sqrt{12}-\sqrt{18}+\sqrt{30}\right)+\sqrt{12}+\sqrt{18}+\sqrt{30}}{12}$

$\sf = \dfrac{-\sqrt{12}+\sqrt{18}-\sqrt{30}+\sqrt{12}+\sqrt{18}+\sqrt{30}}{12}$

$\sf = \dfrac{\sqrt{18}+\sqrt{18}}{12}$

$\sf = \dfrac{2\sqrt{18}}{12}$

$\sf =\dfrac{\sqrt{18}}{6}$

Simplifying the radical.

$\sf =\dfrac{3\sqrt{2}}{6}$

$\sf =\dfrac{\sqrt{2}}{2}$

Taking RHS,

$\sf =\dfrac{1}{\sqrt{2}}$

Rationalising the denominator.

$\sf =\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}$

$\sf =\dfrac{1\left(\sqrt{2}\right)}{\sqrt{2}\sqrt{2}}$

$\sf =\dfrac{\sqrt{2}}{2}$

LHS = RHS

Hence, proved!