Question

Show that : $\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2+b^2}+b}÷\frac{\sqrt{a^2+b^2}-b}{a-\sqrt{a^2-b^2}}=\frac{b^2}{d^2}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2+b^2}+b}÷\frac{\sqrt{a^2+b^2}-b}{a-\sqrt{a^2-b^2}} \\ =\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2+b^2}+b}\times \frac{a-\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-b} \\ =\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}+b}\times \frac{a-\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}-b} \\ =\frac{{\left(a\right)}^2-{\left(\sqrt{a^2-b^2}\right)}^2}{{\left(\sqrt{a^2+b^2}\right)}^2-{\left(b\right)}^2} \\ =\frac{a^2-\left(a^2-b^2\right)}{a^2+b^2-b^2} \\ =\frac{a^2-a^2+b^2}{a^2+b^2-b^2} \\ =\frac{b^2}{a^2} \\ \mathrm{and}\mathrm{RHS}=\frac{b^2}{a^2} \\ \mathrm{LHS}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$