Question

If m=(costheta -sintheta)and n=costheta+sintheta)then prove that √m/n+√n/m=2/√1-tan^2theta​

Answer 1

$\textbf{Given:}$

$m=cos\theta-sin\theta$

$n=cos\theta+sin\theta$

$\displaystyle\frac{m}{n}=\frac{cos\theta-sin\theta}{cos\theta+sin\theta}$

$\text{Divide both numerator and denominator of R.H.S by cos\theta }$

$\text{we get}$

$\displaystyle\frac{m}{n}=\frac{1-tan\theta}{1+tan\theta}$

$\displaystyle\sqrt{\frac{m}{n}}=\sqrt{\frac{1-tan\theta}{1+tan\theta}}$

$\implies\bf\displaystyle\sqrt{\frac{m}{n}}=\frac{\sqrt{1-tan\theta}}{\sqrt{1+tan\theta}}$......(1)

$\text{similarly}$

$\bf\displaystyle\sqrt{\frac{n}{m}}=\frac{\sqrt{1+tan\theta}}{\sqrt{1-tan\theta}}$.......(2)

$\text{Adding (1) and (2), we get}$

$\displaystyle\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{\sqrt{1-tan\theta}}{\sqrt{1+tan\theta}}+\frac{\sqrt{1+tan\theta}}{\sqrt{1-tan\theta}}$

$\displaystyle\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{(\sqrt{1-tan\theta})^2+(\sqrt{1+tan\theta})^2}{\sqrt{1+tan\theta}\sqrt{1-tan\theta}}$

$\displaystyle\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{1-tan\theta+1+tan\theta}{\sqrt{1-tan^2\theta}}$

$\implies\bf\displaystyle\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\frac{2}{\sqrt{1-tan^2\theta}}$