Math, asked by tambolijafar24, 2 months ago

please answer this question

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Answered by Anonymous
0

m^2 - n^2 =

 {a}^{2}  -  \frac{1}{ {a}^{2} }

or

  \frac{ {a}^{4}  - 1}{ {a}^{2} }

Answered by kailashmannem
52

 \Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}

  •  \sf a \: + \: \dfrac{1}{a} \: = \: m \: , \: a \: - \: \dfrac{1}{a} \: = \: n

  • a ≠ 0

 \Large{\bf{\orange{\mathfrak{\dag{\underline{\underline{To \: Find:-}}}}}}}

  • m² - n²

\Large{\bf{\red{\mathfrak{\dag{\underline{\underline{Solution:-}}}}}}}

  • m² - n²

Here,

  •  \sf a \: + \: \dfrac{1}{a} \: = \: m

  •  \sf a \: - \: \dfrac{1}{a} \: = \: n

We know that,

  • a² - b² = (a + b)(a - b)

Here,

  • m² - n² = (m + n)(m - n)

Substituting the values,

  • (m + n)(m - n)

  •  \sf \bigg(a \: + \: \dfrac{1}{a} \: + \: a \: - \: \dfrac{1}{a}\bigg)\bigg(a \: + \: \dfrac{1}{a} \: - \big(a \: - \: \dfrac{1}{a}\big)\bigg)

  •  \sf \bigg(a \: + \: \dfrac{1}{a} \: + \: a \: - \: \dfrac{1}{a}\bigg)\bigg(a \: + \: \dfrac{1}{a} \: - a \: + \: \dfrac{1}{a}\bigg)

  •  \sf \bigg(a \: \cancel{+ \: \dfrac{1}{a}} \: + \: a \: \cancel{- \: \dfrac{1}{a}}\bigg)\bigg(\cancel{a} \: + \: \dfrac{1}{a} \: \cancel{- \: a} \: + \: \dfrac{1}{a}\bigg)

  •  \sf \big(a \: + \: a \big)\bigg(\dfrac{1}{a} \: + \: \dfrac{1}{a}\bigg)

  •  \sf \big(2a \big)\bigg(\dfrac{2}{a} \bigg)

  •  \sf 2a \: * \: \dfrac{2}{a}

  •  \sf 2\cancel{a} \: * \: \dfrac{2}{\cancel{a}}

  •  \sf 2 \: * \: \dfrac{2}{1}

  •  \sf 2 \: * \: 2

  •  \sf 4

Therefore,

  • - = 4
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