sqrt{2} + sqrt{3} + sqrt{4} + sqrt{6} is equa | vedclass

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(A) We know the identity $\cot(A) = \frac{1 + \cos(2A)}{\sin(2A)}$.Let $A = 7.5^{\circ}$,then $2A = 15^{\circ}$.Substituting $A = 7.5^{\circ}$ in the

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$\cot(7.5^{\circ})$

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(A) We know the identity $\cot(A) = \frac{1 + \cos(2A)}{\sin(2A)}$.Let $A = 7.5^{\circ}$,then $2A = 15^{\circ}$.Substituting $A = 7.5^{\circ}$ in the identity,we get $\cot(7.5^{\circ}) = \frac{1 + \cos(15^{\circ})}{\sin(15^{\circ})}$.We know that $\cos(15^{\circ}) = \cos(45^{\circ} - 30^{\circ}) = \cos(45^{\circ})\cos(30^{\circ}) + \sin(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$.Similarly,$\sin(15^{\circ}) = \sin(45^{\circ} - 30^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) - \cos(45^{\circ})\sin(30^{\circ}) = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}}$.Substituting these values: $\cot(7.5^{\circ}) = \frac{1 + \frac{\sqrt{3} + 1}{2\sqrt{2}}}{\frac{\sqrt{3} - 1}{2\sqrt{2}}} = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}$.Rationalizing the denominator: $\frac{(2\sqrt{2} + \sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2\sqrt{6} + 2\sqrt{2} + 3 + \sqrt{3} + \sqrt{3} + 1}{3 - 1} = \frac{2\sqrt{6} + 2\sqrt{2} + 2\sqrt{3} + 4}{2} = \sqrt{6} + \sqrt{2} + \sqrt{3} + 2$.Since $\sqrt{4} = 2$,the expression is $\sqrt{6} + \sqrt{2} + \sqrt{3} + \sqrt{4}$.

$\sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$ is equal to

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