If tan A = root2-1, then …
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Posted by Shyni Manoj 4 years, 7 months ago
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Gaurav Seth 4 years, 7 months ago
Tan A = √2 - 1
Tan A = p / b { p: perpendicular , b:base}
h : hypotenuse
(√ 2 - 1 ) / 1 = p / b
p= √2 - 1 , b = 1
By pythagoras theorem ,
p² + b² = h²
(√2 - 1 )² + (1)² = h²
2 + 1 - 2√2 + 1 = h ²
h² = [4 - 2√2 ]
Sin A = p / h , Cos A = b / h
Sin A Cos A = ( p * b ) / h²
=> (√ 2 - 1 ) / ( 4 - 2 √2)
Rationalize Denomonator ,
=> ( √ 2 - 1 ) * ( 4 + 2 √2 ) / ( 4 - 2 √2 )(4 + 2 √2)
=> (4 √2 + 4 - 4 - 2√2 ) / ( 16 - 8 )
=> 2 √2 / 8
=> √2 / 4
Hence Proved
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