Question

find a and b if root 2 +1 /root 2-1 -root2-1/root 2+1=a+toot2b

Answer 1

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 2

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