Find the value of a and b in -2 - 1-2 - 1 - -2 - 1-2 - 1 - a - -2 b-

Given, √(2) #160;+ 1√(2) #160; 1 √(2) #160; 1√(2) #160;+ 1 = a + √(2) b (√(2) #160;+ 1)^2 (√(2) 1)^2(√(2))^2 #160; 1^2= a + √(2) b ...

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Rationalisation

Find the valu...

Question

Find the value of $a$ and $b$ in $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = a + \sqrt{2} b$ .

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a

b,

\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = a + \sqrt{2} b

a^{2} b^{2} = 1

S

Δ

Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

\frac{4 S}{3}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + a^{2} - b^{2} & 2 a b & - 2 b 2 a b & 1 - a^{2} + b^{2} & 2 a 2 b & - 2 a & 1 - a^{2} - b^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

(\frac{1}{2}) a + (\frac{1}{2}) b = 3

a + 2 b = \frac{2}{3}

{sin}^{- 1} \frac{2 a}{1 + a^{2}} + {sin}^{- 1} \frac{2 b}{1 + b^{2}} = 2 {tan}^{- 1} x

x

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