Question

If $m=(cos\theta -sin\theta )$ and $n=(cos\theta +sin\theta )$ then show that

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

Answer 1

$\sqrt{\frac{m}{n}}+\sqrt{\frac{n}{m}}=\sqrt{\frac{(cos\theta -sin\theta )}{(cos\theta +sin\theta }}+\sqrt{\frac{(cos\theta +sin\theta }{(cos\theta -sin\theta )}}$

=$\sqrt{\frac{(cos\theta -sin\theta )}{(cos\theta +sin\theta }}\times \sqrt{\frac{(cos\theta -sin\theta }{(cos\theta -sin\theta )}}+\sqrt{\frac{(cos\theta +sin\theta }{(cos\theta -sin\theta )}}\times \sqrt{\frac{(cos\theta +sin\theta )}{(cos\theta +sin\theta }}$

=$\frac{(cos\theta -sin\theta )}{\sqrt{co{s}^2\theta -si{n}^2\theta }}+\frac{cos\theta +sin\theta }{\sqrt{co{s}^2\theta -si{n}^2\theta }}$

=$\frac{\left(2\cos \theta \right)}{\sqrt{co{s}^2\theta -si{n}^2\theta }}$

=$\frac{2\sqrt{co{s}^2\theta }}{\sqrt{co{s}^2\theta -si{n}^2\theta }}$

=$\frac{2}{\sqrt{1-ta{n}^2\theta }}$