Question

if a =√2+1 , find the value of (a-1/a2)2​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\implies ( {a - \frac{1}{a}})^{2}=1}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies a = \sqrt{2} + 1 \\ \\ \underline \bold{to \: find: } \\ \implies ( {a - \frac{1}{a} })^{2} = ?$

• According to given question :

$\bold{Using \: identity \to } \\ \bold{ ({a - b})^{2} = {a}^{2} + {b}^{2} - 2ab} \\ \\ \bold{For \: finding \: values : } \\ \implies ( {a - \frac{1}{a} })^{2} = ({a})^{2} + (\frac{1}{a})^{2} - 2 \times \cancel{a} \times \frac{1}{ \cancel{a}} \\ \\ \implies ( {a - \frac{1}{a} })^{2} = ({ \sqrt{2} + 1})^{2} + (\frac{1}{ \sqrt{2} + 1 })^{2} - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = ({ \sqrt{2} })^{2} + {1}^{2} + 2 \sqrt{2} + \frac{1}{ \sqrt{2}^{2} + {1}^{2} + 2 \sqrt{2} } - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = 2 + 1 + 2 \sqrt{2} + \frac{1}{2 + 1 + 2 \sqrt{2} } - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = 3 + 2 \sqrt{3} + \frac{1}{3 + 2\sqrt{2} } \times \frac{3 - 2\sqrt{2} }{3 - 2 \sqrt{2} } - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = 3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2} }{9 - 4 \times 2} - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = 3 + 2 \sqrt{2} + \frac{3 - 2 \sqrt{2} }{9 - 8} - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = 3 + \cancel{2 \sqrt{2}} + 3 \cancel{- 2 \sqrt{2}} - 2 \\ \\ \implies ( {a - \frac{1}{a} })^{2} = 3 - 2 \\ \\ \bold{\implies ( {a - \frac{1}{a} })^{2} = 1}$

Answer 2

Answer:

(a-1/a2)2=1

Step-by-step explanation:

if a =√2+1 , find the value of (a-1/a2)2