Question

Solve the question given in attachment...

Ans :- (b) 1​

Answer 1

Answer:

Option(B)

Step-by-step explanation:

$Given:\frac{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1)}(\sqrt{3}-\sqrt{2})}{(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1})}$

On rationalizing, we get

$=\frac{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})(\sqrt{3}-\sqrt{2})}{(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1})}*\frac{(\sqrt{\sqrt{3}+1} +\sqrt{\sqrt{3}-1})}{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})}$

$=\frac{(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})(\sqrt{3}-\sqrt{2})(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})}{(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1})(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}-1})}$

$=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{\sqrt{3}+1}+\sqrt{\sqrt{3}}-1)^2}{(\sqrt{\sqrt{3} + 1)^2}-\sqrt{\sqrt{3}-1)}^2}$

$=\frac{(\sqrt{3}-\sqrt{2})(2\sqrt{3}+2\sqrt{2})}{\sqrt{3}+1-\sqrt{3}+1}$

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

$=\frac{6-4}{2}$

$=\frac{2}{2}$

$=1$

Hope it helps!

Answer 2

Note: Check this attachment.