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24 votes
Prove that:-


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


User Andrej Kesely
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1 Answer

25 votes
25 votes

Answer:


\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. }

User Nilo Alan
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