Question

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

Answer 1

Given that, $tanx=\frac{b}{a}$

$\therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{\sqrt{(a+b{)}^2}+\sqrt{(a-b{)}^2}}{\sqrt{(a-b)}\sqrt{(a+b)}}=\frac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}=\frac{2a}{\sqrt{a^2-b^2}}=\frac{2aa}{a\sqrt{1-{\left(\frac{b}{a}\right)}^2}}[\because \frac{b}{a}=tanx]=\frac{2}{\sqrt{1-ta{n}^{2}x}}=\frac{2cosx}{\sqrt{co{s}^{2}x-si{n}^{2}x}}$

$[\because cos2x=co{s}^{2}x-si{n}^{2}x]=\frac{2cosx}{\sqrt{cos2x}}$