# tan

Question:

$\tan 82 \frac{1^{\circ}}{2}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$

Solution:

Here,

$\tan (82.5)^{\circ}=\tan (90-7.5)^{\circ}$

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

$=\frac{1}{\tan (7.5)^{\circ}}$

We know,

$\tan \left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}$

On putting $x=15^{\circ}$, we get

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

$=\frac{\sin (45-30)^{\circ}}{1+\cos (45-30)^{\circ}}$

$=\frac{\sin 45^{\circ} \cos 30^{\circ}-\sin 30^{\circ} \cos 45^{\circ}}{1+\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}}$

$=\frac{\left(\frac{1}{\sqrt{2}}\right) \times\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right) \times\left(\frac{1}{\sqrt{2}}\right)}{1+\left(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}\right)+\left(\frac{1}{\sqrt{2}} \times \frac{1}{2}\right)}$

$=\frac{\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}}{1+\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}}$

$=\frac{\sqrt{3}-1}{2 \sqrt{2}+\sqrt{3}+1}$

Now,

$\tan (82.5)^{\circ}=\frac{1}{\tan (7.5)^{\circ}}$

$=\frac{2 \sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

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

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

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

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

$=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$                ...(1)

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

$=\sqrt{3}(\sqrt{2}+1)+\sqrt{2}(\sqrt{2}+1)$

$=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)$              ..(2)

From eqs. $(1)$ and $(2)$, we get

$\tan (82.5)^{\circ}=(\sqrt{3}+\sqrt{2})(\sqrt{2}+1)=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$