Question

Prove that:

$tan82{\frac{1}{2}}^{\circ }=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

Answer 1

$tan82{\frac{1}{2}}^{\circ }=tan{(90-7\frac{1}{2})}^{\circ }=cot7{\frac{1}{2}}^{\circ }=cotA\text{If}A=7\frac{1}{2}\circ$

Now

$cotA=\frac{cosA}{sinA}=\frac{2co{s}^{2}A}{2sinAcosA}=\frac{1+co{s}^{2}A}{si{n}^{2}A}cotA=\frac{1+cos15}{sin15}=\frac{1+cos(45-30)}{sin15}$

$=\frac{1+(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2})}{\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}}=\frac{2\sqrt{2}+(\sqrt{3}+1)}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2\sqrt{2}(\sqrt{3}+1)+(\sqrt{3}+1{)}^2}{3-1}=\frac{2\sqrt{6}+2\sqrt{2}+4+2\sqrt{3}}{2}cotA=\sqrt{6}+\sqrt{2}+2+\sqrt{3}\cdots \left(i\right)=\sqrt{2}+2+\sqrt{6}+\sqrt{3}$

$=\sqrt{2}+(1+\sqrt{2})+\sqrt{3}(\sqrt{2}+1)cotA=(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})\cdots \left(ii\right)$

From equation (i) and (ii)

$tan82{\frac{1}{2}}^{\circ }=cot7{\frac{1}{2}}^{\circ }=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}=(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})$