Question

If $x=\frac{2ab}{a+b}$ then find the value of $\frac{x+a}{x-a}+\frac{a+b}{a-b}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ x=\frac{2ab}{a+b} \\ \therefore \frac{x}{a}=\frac{2b}{a+b},\frac{x}{b}=\frac{2a}{a+b}...\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{x+a}{x-a}+\frac{x+b}{x-b}. \\ \mathrm{On}\mathrm{applying}\mathrm{componendo}\&\mathrm{dividendo},\mathrm{we}\mathrm{get}: \\ \frac{x+a+x-a}{x+a-x+a}+\frac{x+b+x-b}{x+b-x+b} \\ =\frac{2x}{2a}+\frac{2x}{2b} \\ =\frac{x}{a}+\frac{x}{b} \\ \mathrm{From}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \frac{2b}{a+b}+\frac{2a}{a+b} \\ =\frac{2\left(b+a\right)}{a+b} \\ =2 \end{gathered}$$