find a and b if root 2 +1 /root 2-1 -root2-1/root 2+1=a+toot2b
Answers
Answer:
(0,4)
Step-by-step explanation:
Given Find a and b if root 2 +1 /root 2-1 -root2-1/root 2+1=a+toot2b
√2 + 1 / √2 - 1 - √2 - 1 / √2 + 1 = a + √b
Taking lcm we get
(√2 + 1)^2 - (√2 - 1)^2 /(√2 -1)(√2 + 1)
(√2)^2 + 1 + 2√2 - ((√2)^2 + 1 - 2√2) / 2 - 1
2 + 1 + 2√2 - (2 + 1 - 2√2)
a + √2 b = 4√2
√2 b = 4√2
b = 4, a = 0
So a = 0, b = 4
Answer:
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