Given-m-cos-sin-n-cos-sin-then-m-n-n-m-

Question Number 144727 by imjagoll last updated on 28/Jun/21

Given {\begin{cases} m = \cos θ - \sin θ \\ n = \cos θ + \sin θ \end{cases}

then \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = ?

Answered by liberty last updated on 28/Jun/21

\Leftrightarrow \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{m + n}{\sqrt{mn}}

{\begin{cases} m^{2} = \cos^{2} - \sin 2 θ + \sin^{2} θ \\ n^{2} = \cos^{2} θ + \sin 2 θ + \sin^{2} θ \end{cases}

{\begin{cases} m^{2} = 1 - \sin 2 θ \\ n^{2} = 1 + \sin 2 θ \end{cases} \Leftrightarrow m^{2} + n^{2} = 2

\Rightarrow {(m + n)}^{2} - 2 mn = 2

\Rightarrow m + n = \sqrt{2 + 2 mn}

\Rightarrow m . n = \cos^{2} θ - \sin^{2} θ = \cos 2 θ

then \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{\sqrt{2 + 2 \cos 2 θ}}{\sqrt{\cos 2 θ}}

= \sqrt{2 \sec 2 θ + 2}

Commented by Rasheed.Sindhi last updated on 28/Jun/21

T y p o s i r!

{\begin{cases} m^{2} = 1 - \sin 2 θ \\ n^{2} = 1 + \sin 2 θ \end{cases} \Leftrightarrow m^{2} + n^{2} = 2

Commented by liberty last updated on 28/Jun/21

yes sir . thank you

Answered by Olaf_Thorendsen last updated on 28/Jun/21

m = \cos θ - \sin θ

n = \cos θ + \sin θ

x = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}}

x^{2} = \frac{m}{n} + 2 + \frac{n}{m}

x^{2} = \frac{m^{2} + n^{2}}{m n} + 2

x^{2} = \frac{(1 - \sin 2 θ) + (1 + \sin 2 θ)}{\cos^{2} θ - \sin^{2} θ} + 2

x^{2} = \frac{2 + 2 \cos 2 θ}{\cos 2 θ}

x^{2} = \frac{{4 \cos}^{2} θ}{\cos 2 θ}

x = \frac{2 ∣ \cos θ ∣}{\sqrt{\cos 2 θ}}