Question

Prove:
a^-1/a^-1+b^-1 + a^-1/a^-1-b^-1 = 2b^2/b^2-a^2

Answer 1

Answer:

$\frac{ {a}^{ - 1} }{ {a}^{ - 1} + {b}^{ - 1} } + \frac{ {a}^{ - 1} }{ {a}^{ - 1} - {b}^{ - 1} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } \\ \\ \frac{ \frac{1}{a} }{ \frac{1}{a} + \frac{1}{b} } + \frac{ \frac{1}{a} }{ \frac{1}{a} - \frac{1}{b} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } \\ \\ \frac{ \frac{1}{a} }{ \frac{b + a}{ab} } + \frac{ \frac{1}{a} }{ \frac{b - a}{ab} } = \frac{2 {b}^{2} }{ {b }^{2} - {a}^{2} } \\ \\ \frac{1}{a} \times \frac{ab}{b + a} + \frac{1}{a} \times \frac{ab}{b - a} = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } \\ \\ \frac{b}{b + a} + \frac{b}{b - a} = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } \\ \\ \frac{b( b - a) + b(b + a)}{(b + a)(b - a)} = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } \\ \\ \frac{b {}^{2} - ab + {b}^{2} + ab }{ {b}^{2} - {a}^{2} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } \\ \\ \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} } = \frac{2 {b}^{2} }{ {b}^{2} - {a}^{2} }$

Therefore, LHS=RHS.

Note:

(a+b)(a-b)=a²-b²