Question

State true or false:

$\left[\frac{\tan {60}^{\circ }+1}{\tan {60}^{\circ }-1}\right]=\frac{1+\tan {30}^{\circ }}{1-\tan {30}^{\circ }}$

• A. True
• B. False

Answer 1

The correct option is A True

$\left[\frac{\tan {60}^{\circ }+1}{\tan {60}^{\circ }-1}\right]=\frac{1+\tan {30}^{\circ }}{1-\tan {30}^{\circ }}$
L.H.S: $\left[\frac{\tan {60}^{\circ }+1}{\tan {60}^{\circ }-1}\right]$

$=\left[\frac{\sqrt{3}+1}{\sqrt{3}-1}\right]$

$=\left[\frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\right]$

$=\left[\frac{3+1+2\sqrt{3}}{3-1}\right]$

$=\left[\frac{4+2\sqrt{3}}{2}\right]$

$=[2+\sqrt{3}]$
R.H.S: $\frac{1+\tan {30}^{\circ }}{1-\tan {30}^{\circ }}$

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

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

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

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

= $2+\sqrt{3}$