$\frac{4}{1+\sqrt{2}+\sqrt{3}}$ is equal to
$sqrt{2}-sqrt{3}+2$
$sqrt{2}+sqrt{3}+2$
$sqrt{2}+sqrt{6}+2$
$sqrt{2}-sqrt{6}+2$
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2024
We can group the terms in the denominator, for instance, as $(1+\sqrt{2}) + \sqrt{3}$. We multiply the numerator and denominator by the conjugate of this expression, which is $(1+\sqrt{2}) – \sqrt{3}$.
Expression = $\frac{4}{(1+\sqrt{2})+\sqrt{3}} \times \frac{(1+\sqrt{2})-\sqrt{3}}{(1+\sqrt{2})-\sqrt{3}}$
Numerator = $4(1+\sqrt{2}-\sqrt{3})$.
Denominator = $((1+\sqrt{2})+\sqrt{3})((1+\sqrt{2})-\sqrt{3})$. This is in the form $(a+b)(a-b) = a^2 – b^2$, where $a = (1+\sqrt{2})$ and $b = \sqrt{3}$.
Denominator = $(1+\sqrt{2})^2 – (\sqrt{3})^2$.
$(1+\sqrt{2})^2 = 1^2 + 2(1)(\sqrt{2}) + (\sqrt{2})^2 = 1 + 2\sqrt{2} + 2 = 3 + 2\sqrt{2}$.
$(\sqrt{3})^2 = 3$.
Denominator = $(3 + 2\sqrt{2}) – 3 = 2\sqrt{2}$.
So the expression becomes $\frac{4(1+\sqrt{2}-\sqrt{3})}{2\sqrt{2}}$.
We can simplify by dividing 4 by 2: $\frac{2(1+\sqrt{2}-\sqrt{3})}{\sqrt{2}}$.
Now, we need to rationalize the denominator $\sqrt{2}$ by multiplying the numerator and denominator by $\frac{\sqrt{2}}{\sqrt{2}}$.
Expression = $\frac{2(1+\sqrt{2}-\sqrt{3})}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
Expression = $\frac{2\sqrt{2}(1+\sqrt{2}-\sqrt{3})}{2}$.
Cancel the 2 in the numerator and denominator:
Expression = $\sqrt{2}(1+\sqrt{2}-\sqrt{3})$.
Now, distribute $\sqrt{2}$:
Expression = $\sqrt{2} \times 1 + \sqrt{2} \times \sqrt{2} – \sqrt{2} \times \sqrt{3}$
Expression = $\sqrt{2} + 2 – \sqrt{6}$.
Rearranging the terms to match the options, we get $2 + \sqrt{2} – \sqrt{6}$, which is the same as $\sqrt{2}-\sqrt{6}+2$.