Question

Prove that, cot $7{\frac{1}{2}}^{\circ }$ or tan $82{\frac{1}{2}}^{\circ }$ = $(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)$ or $\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

Answer 1

We need to find value of $\cot \left(7\frac{1}{2}\right)$

$\cot \theta =\frac{\cos \theta }{\sin \theta }$

$\sin \left({15}^o\right)=\sin (45-{30}^o)$

$=\sin {45}^o\cos {30}^o-\cos {45}^o\sin {30}^o$

or $\sin (A-B)=\sin A\cos B-\cos A\cos B$(Trigonometric property)

$=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)=\frac{1}{2\sqrt{2}}(\sqrt{3}-1)=\sin {15}^o)$

Similarly $\cos {15}^o=\cos (45-{30}^o)$

$\cos {45}^o\cos {30}^o+\sin {45}^o\sin 30$

as $\cos (A-B)=\cos A\cos B+\sin A\sin B$

$=\left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{2}}{2}\right)+\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}=\cos 15$

$\cot \left(7\frac{1}{2}\right)=\frac{\cos \left(7\frac{1}{2}\right)}{\sin \left(7\frac{1}{2}\right)}=\frac{2\cos \left(7\frac{1}{2}\right)\cos \left(7\frac{1}{2}\right)}{2\sin \left(7\frac{1}{x}\right)\cos \left(\frac{1}{2}\right)}=\frac{2{\cos }^2\left(7\frac{1}{2}\right)}{\sin (2\times 7\frac{1}{2})}$

(multiply and divide by $2\cos \left(7\frac{1}{2}\right)$)

as $\sin \left(2A\right)-2\sin A\cos A,\cos 2A=2{\cos }^{2}A-1$

So, $\cot \left(7\frac{1}{2}\right)=\frac{1+\cos {15}^o}{\sin {15}^o}=\frac{1+\frac{\sqrt{3}+1}{2\sqrt{2}}}{\frac{\sqrt{3}-1}{2\sqrt{2}}}=\frac{1+\sqrt{3}+2\sqrt{2}}{\sqrt{3}-1}$

$=\frac{(1+\sqrt{3}+2\sqrt{2})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{1+\sqrt{3}+2\sqrt{2}+\sqrt{3}+3+2\sqrt{6}}{(\sqrt{3}{)}^2-(1{)}^2}$

So, $\cot \left(7\frac{1}{2}\right)=(\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6})=(\sqrt{2}+\sqrt{3})(\sqrt{2}+1)$