Question

$\large\begin{array}{l}\mathsf{Calcular~cos(x)~sabendo~que~cotg(x)=\dfrac{2\sqrt{m}}{m-1}~~com~m\ \textgreater \ 1.}\end{array}$



$\mathsf{gab:~cos(x)=\pm~\dfrac{2 \sqrt{m} }{m+1} }$

Answer 1

$cotg(x) = \frac{1}{tg(x)} = \frac{cos(x)}{sen(x)}\\ \\ cotg(x) = \frac{2\sqrt{m}}{m-1}\\ \\ \frac{cos(x)}{sen(x)} = \frac{2\sqrt{m}}{m-1}\\ \\ Lembrando~que:\\ \\ \boxed{sen^2(x)+cos^2(x)=1}\\ \\ sen(x) = \frac{cos(x)(m-1)}{2\sqrt{m})}\\ \\ (\frac{cos(x)(m-1)}{2\sqrt{m})})^2+cos^2(x) = 1\\ \\ \frac{cos^2(x)(m-1)^2}{4m}+cos^2(x) = 1\\ \\ \frac{cos^2(x)(m-1)^2+4m~\cdot~cos^2(x)}{4m} = 1\\ \\ cos^2(x)(m^2-2m+1+4m) = 4m\\ \\ cos^2(x)(m^2+2m+1) = 4m$

$cos^2(x) = \frac{4m}{(m+1)^2}\\ \\ cos(x) = \pm\sqrt{\frac{4m}{(m+1)^2}}\\ \\ \large\boxed{cos(x) = \pm\frac{2\sqrt{m}}{m+1}}}$