Question

$\sqrt{1 + cos 30} \div \sqrt{1 - cos 30} - tan60 = cosec30$
Prove ​

Answer 1

Step-by-step explanation:

THE ANSWER OF UR QUESTION...

$\cos( \sqrt{30 + 1 = 901....( \sqrt{ \cos(30 - 1 = 899) } } ) .... \\ \tan(60 \div 2 \cos( = 30 \csc(?) ) )$

I M IN STANDARD 8 SO..IT MAY CORRECT OR NOT CORRECT.

Answer 2

Answer:

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Step-by-step explanation:

$to \: prove = \\ \\ \frac{ \sqrt{1 + cos \: 30} }{ \sqrt{1 - cos \: 30} } \: - tan \: 60 = cosec \: 30 \\ \\ le t\: then \\ \\ LHS = \frac{ \sqrt{1 + cos \: 30} }{ \sqrt{1 - cos \: 30} } \: - tan \: 60 \\ \\ RHS = cosec \: 30$

$considering \: LHS \\ we \: know \: that \\ \\ cos \: 30 = \frac{ \sqrt{3} }{2} \\ \\ thus \: then \\ \\ = \frac{ \sqrt{1 + cos \: 30} }{ \sqrt{1 - cos \: 30} } \: - tan \: 60 \\ \\ = \frac{ \sqrt{1 + \frac{ \sqrt{3} }{2} } }{ \sqrt{1 - \frac{ \sqrt{3} }{2} } } \: - tan \: 60 \\ \ \\ = \frac{ \frac{ \frac{ \sqrt{2 + \sqrt{3} } }{ \sqrt{2} } }{ \sqrt{2 - \sqrt{3} } } }{ \sqrt{2} } \: - tan \: 60 \\ \\ = \frac{ \sqrt{2 + \sqrt{3} } }{ \sqrt{2 - \sqrt{3} } } \: - tan \: 60 \\ \\ so \: rationalising \: \: denominator \\ we \: get \\ \\ = \frac{ \sqrt{2 + \sqrt{3} } }{ \sqrt{2 - \sqrt{3} } } \times \frac{ \sqrt{2 + \sqrt{3} } }{ \sqrt{2 + \sqrt{3} } } \: - tan \: 60 \\ \\ = \frac{( \sqrt{2 + \sqrt{3} } \: ) {}^{2} }{( \sqrt{2 - \sqrt{3} })( \sqrt{2 + \sqrt{3} }) {} } - tan \: 60\\ \\ = \frac{2 + \sqrt{3} }{(2 - \sqrt{3} )(2 + \sqrt{3} )} - \: tan \: 60 \\ \\ \\ = \frac{2 + \sqrt{3} }{(2) {}^{2} - ( \sqrt{3} ) {}^{2} } \: - tan \: 60 \\ \\ = \frac{2 + \sqrt{3} }{4 - 3} \: - tan \: 60 \\ \\ = \frac{2 + \sqrt{3} }{1} \: - tan \: 60 \\ \\ = 2 + \sqrt{3} - tan \: 60 \\ \\ we \: have \\ tan \: 60 = \sqrt{3} \\ \\ thus \: then \\ \\ LHS= 2$

$now \: considering \: \: RHS \\ we \: know \: that \\ \\ cosec \: 30 = 2 \\ \\ thus \: then \: \\ RHS = 2$

$hence \: then \\ LHS=RHS \\ \\ thus \: proved$