Question

Prove that:-


`\sqrt{ \frac{1 + cos30 {}^( \circ) }{1 - cos {30}^( \circ) } } = sec \: {60}^( \circ) + tan \: {60}^( \circ)`
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Answer 1

`\displaystyle{ \sqrt{ \frac{1 + cos \: {30}^( \circ) }{1 - cos \: {30}^( \circ) } } = sec \: {60}^( \circ) + tan \: {60}^( \circ) }`

`LHS = \displaystyle{ \sqrt{ \frac{1 + cos \: {30}^( \circ) }{1 - cos \: {30}^( \circ) } } }`

`\displaystyle{ \sqrt{ (1 + ( √(3) )/(2) )/(1 - ( √(3) )/(2) ) } }`

`\displaystyle{ \sqrt{ ( (2 + √(3) )/(2) )/( (2 - √(3) )/(2) ) } }`

`\displaystyle{ \sqrt{ (2 + √(3) )/(2 - √(3) ) * (2 + √(3) )/(2 + √(3) ) } }`

`\displaystyle{ \sqrt{ \frac{(2 + √(3) {)}^(2) }{4 - 3} } }`

`\displaystyle{2 + √(3) }`

`RHS = sec \: {60}^( \circ) + tan \: {60}^( \circ)`

`= 2 + √(3)`

`\rm\therefore{LHS=RHS,proved. }`