2021-08-20

If $\mathrm{tan}x=\frac{b}{a}$, then find the value of $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}$

Mayme

Solution:
$\mathrm{tan}x=\frac{b}{a}$
$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\sqrt{\frac{a\left(1+\frac{b}{a}\right)}{a\left(1-\frac{b}{a}\right)}}+\sqrt{\frac{a\left(1-\frac{b}{a}\right)}{a\left(1+\frac{b}{a}\right)}}$
$=\sqrt{\frac{\left(1+\frac{b}{a}\right)}{\left(1-\frac{b}{a}\right)}}+\sqrt{\frac{\left(1-\frac{b}{a}\right)}{\left(1+\frac{b}{a}\right)}}$
$=\sqrt{\frac{1+\mathrm{tan}x}{1-\mathrm{tan}x}}+\sqrt{\frac{1-\mathrm{tan}x}{1+\mathrm{tan}x}}$
$=\frac{{\left(\sqrt{1+\mathrm{tan}x}\right)}^{2}+{\left(\sqrt{1-\mathrm{tan}x}\right)}^{2}}{\sqrt{1-\mathrm{tan}x}\sqrt{1+\mathrm{tan}x}}$
$=\frac{\left(1+\mathrm{tan}x\right)+\left(1-\mathrm{tan}x\right)}{\sqrt{\left(1-\mathrm{tan}x\right)\left(1+\mathrm{tan}x\right)}}$
$=\frac{2}{\sqrt{1-{\mathrm{tan}}^{2}x}}$
$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{2}{\sqrt{1-\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}}$
$=\frac{2}{\sqrt{\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}x}}}$
$=\frac{2}{\frac{\sqrt{\mathrm{cos}2x}\left\{}{\mathrm{cos}x}}$
$\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{2\mathrm{cos}x}{\sqrt{{\mathrm{cos}}^{2}x}}$

Jeffrey Jordon