Question

The value of ${\left(\frac{\tan {60}^0+1}{\tan {60}^0-1}\right)}^2=\frac{1+\cos {30}^0}{1-\cos {30}^0}$

if true then enter $1$ and if false then write $0$

Answer 1

${\left(\frac{\tan {60}^0+1}{\tan {60}^0-1}\right)}^2=\frac{1+\cos {30}^0}{1-\cos {30}^0}$
L.H.S: ${\left(\frac{\tan {60}^0+1}{\tan {60}^0-1}\right)}^2$

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

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

= $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
R.H.S: $\frac{1+\cos {30}^0}{1-\cos {30}^0}$

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

= $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Thus, $L.H.S=R.H.S$