Answer:
[tex] \displaystyle{ \sqrt{ \frac{1 + cos \: {30}^{ \circ} }{1 - cos \: {30}^{ \circ} } } = sec \: {60}^{ \circ} + tan \: {60}^{ \circ} }[/tex]
[tex]LHS = \displaystyle{ \sqrt{ \frac{1 + cos \: {30}^{ \circ} }{1 - cos \: {30}^{ \circ} } } }[/tex]
[tex] \displaystyle{ \sqrt{ \frac{1 + \frac{ \sqrt{3} }{2} }{1 - \frac{ \sqrt{3} }{2} } } }[/tex]
[tex] \displaystyle{ \sqrt{ \frac{ \frac{2 + \sqrt{3} }{2} }{ \frac{2 - \sqrt{3} }{2} } } }[/tex]
[tex] \displaystyle{ \sqrt{ \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } } }[/tex]
[tex] \displaystyle{ \sqrt{ \frac{(2 + \sqrt{3} {)}^{2} }{4 - 3} } }[/tex]
[tex] \displaystyle{2 + \sqrt{3} }[/tex]
[tex]RHS = sec \: {60}^{ \circ} + tan \: {60}^{ \circ} [/tex]
[tex] = 2 + \sqrt{3} [/tex]
[tex] \rm\therefore{LHS=RHS,proved. }[/tex]
