Question

Prove that: $cot7\frac{1^{\circ }}{2}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

Answer 1

$\cot 7{\frac{1}{2}}^{\circ }=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

$\Rightarrow L.H.S=\cot 7{\frac{1}{2}}^{\circ }=\cot {\frac{15}{2}}^{\circ }$

$=\frac{\cos {\left(\frac{15}{2}\right)}^{\circ }}{\sin {\left(\frac{15}{2}\right)}^{\circ }}=\frac{2\cos {\left(\frac{15}{2}\right)}^{\circ }}{2\sin {\left(\frac{15}{2}\right)}^{\circ }}$

$=\frac{2{\cos }^2{\left(\frac{15}{2}\right)}^{\circ }}{2\sin {\left(\frac{15}{2}\right)}^{\circ }\cos {\left(\frac{15}{2}\right)}^{\circ }}=\frac{1+\cos 15°}{\sin 15°}$

$[\because 1+\cos 2\theta =2{\cos }^2\frac{2\theta }{2}]$

$[\&2\sin \frac{\theta }{2}.\cos \frac{\theta }{2}=\sin \frac{2\theta }{2}]$

$\Rightarrow \cos 15°=\frac{1}{4}(\sqrt{6}+\sqrt{2})\&\sin 15°=\frac{1}{4}(\sqrt{6}-\sqrt{2})$

$\therefore LHS=\frac{1+\frac{1}{4}(\sqrt{6}+\sqrt{2})}{\frac{1}{4}(\sqrt{6}-\sqrt{2})}=\frac{4+\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

Apply componendo and dividendo, we get

$\Rightarrow LHS=\frac{(4+\sqrt{6}+\sqrt{2})(\sqrt{6}+\sqrt{2})}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})}$

$=\frac{4\sqrt{6}+4\sqrt{2}+6+\sqrt{12}+\sqrt{12}+2}{6-2}$

$=\frac{4\sqrt{6}+4\sqrt{2}+8+2\sqrt{3}+2\sqrt{3}}{4}$

$=\frac{8}{4}+\frac{4\sqrt{2}}{4}+\frac{4\sqrt{3}}{4}+\frac{4\sqrt{6}}{4}$

$=2+\sqrt{2}+\sqrt{3}+\sqrt{6}$

$=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

$\Rightarrow$ $R.H.S$

Hence, the answer is prove.