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Mathematics

If a+1a=ma + \dfrac{1}{a} = m and a ≠ 0; find in terms of 'm'; the value of :

(i) a−1aa - \dfrac{1}{a}

(ii) a2−1a2a^2 - \dfrac{1}{a^2}

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Answer

(i) By formula,

⇒(a+1a)2−(a−1a)2=4\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4

Substituting values we get :

⇒m2−(a−1a)2=4⇒(a−1a)2=m2−4⇒a−1a=±m2−4.\Rightarrow m^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)^2 = m^2 - 4 \\[1em] \Rightarrow a - \dfrac{1}{a} = \pm \sqrt{m^2 - 4}.

Hence, a−1a=±m2−4.a - \dfrac{1}{a} = \pm \sqrt{m^2 - 4}.

(ii) By formula,

⇒a2−1a2=(a−1a)(a+1a)\Rightarrow a^2 - \dfrac{1}{a^2} = \Big(a - \dfrac{1}{a}\Big)\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒⇒a2−1a2=±m2−4×m=±mm2−4.\Rightarrow \Rightarrow a^2 - \dfrac{1}{a^2} = \pm \sqrt{m^2 - 4} \times m \\[1em] = \pm m\sqrt{m^2 - 4}.

Hence, a2−1a2=±mm2−4.a^2 - \dfrac{1}{a^2} = \pm m\sqrt{m^2 - 4}.

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