Question

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

Answer 1

Given that, tan $x=\frac{b}{a}$

$\therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\frac{\sqrt{(a+b{)}^2}+\sqrt{(a-b{)}^2}}{\sqrt{(a-b)}\sqrt{(a+b)}}$

$=\frac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}=\frac{2a}{\sqrt{a^2-b^2}}=\frac{2a}{a\sqrt{1-(\frac{b}{a}{)}^2}}[\because \frac{b}{a}=tanx]$

$=\frac{2}{\sqrt{1-ta{n}^{2}x}}=\frac{2cosx}{\sqrt{co{s}^{2}x-si{n}^{2}x}}$