Question

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

Answer 1

$tan82{\frac{1}{2}}^{\circ }=tan({90}^{\circ }-7{\frac{1}{2}}^{\circ })=cot7{\frac{1}{2}}^{\circ }=cotA$ say where A = $7{\frac{1}{2}}^{\circ }$
Now $cotA=\frac{cosA}{sinA}=\frac{2co{s}^{2}A}{2sinAcosA}$
$=\frac{1+cos2A}{sin2A}=\frac{1+cos{15}^{\circ }}{sin{15}^{\circ }}$
Now put the values of $cos{15}^{\circ }$ and $sin{15}^{\circ }$ from
R.(33) P.518
$\therefore cot7\frac{1^{\circ }}{2}=\frac{2\sqrt{2}+(\sqrt{3}+1)}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$
$=\frac{2\sqrt{2}(\sqrt{3}+1)(\sqrt{3}+1{)}^2}{3-1}$
$=\frac{2\sqrt{6}+2\sqrt{2}+4+2\sqrt{3}}{2}$
$=\sqrt{6}+\sqrt{2}+2+\sqrt{3}$
$=\sqrt{2}(\sqrt{2}+1)+\sqrt{3}(\sqrt{2}+1)$
$(\sqrt{2}+1)\left(\sqrt{3}+\sqrt{2}\right)$