KnowledgeBoat Logo
|

Mathematics

Prove that :

a−1a−1+b−1+a−1a−1−b−1=2b2b2−a2\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}

Indices

6 Likes

Answer

To prove:

a−1a−1+b−1+a−1a−1−b−1=2b2b2−a2\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}

Solving L.H.S. of the above equation, we get :

⇒a−1a−1+b−1+a−1a−1−b−1=1a1a+1b+1a1a−1b=1ab+aab+1ab−aab=aba(b+a)+aba(b−a)=bb+a+bb−a=b(b−a)+b(b+a)(b+a)(b−a)=b2−ba+b2+abb2−a2=2b2b2−a2.\Rightarrow \dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} + \dfrac{1}{b}} + \dfrac{\dfrac{1}{a}}{\dfrac{1}{a} - \dfrac{1}{b}} \\[1em] = \dfrac{\dfrac{1}{a}}{\dfrac{b + a}{ab}} + \dfrac{\dfrac{1}{a}}{\dfrac{b - a}{ab}} \\[1em] = \dfrac{ab}{a(b + a)} + \dfrac{ab}{a(b - a)} \\[1em] = \dfrac{b}{b + a} + \dfrac{b}{b - a} \\[1em] = \dfrac{b(b - a) + b(b + a)}{(b + a)(b - a)} \\[1em] = \dfrac{b^2 - ba + b^2 + ab}{b^2 - a^2} \\[1em] = \dfrac{2b^2}{b^2 - a^2}.

Since, L.H.S. = R.H.S. = 2b2b2−a2\dfrac{2b^2}{b^2 - a^2}

Hence, proved that a−1a−1+b−1+a−1a−1−b−1=2b2b2−a2\dfrac{a^{-1}}{a^{-1} + b^{-1}} + \dfrac{a^{-1}}{a^{-1} - b^{-1}} = \dfrac{2b^2}{b^2 - a^2}.

Answered By

3 Likes

Related Questions