[tex]\bf sec(\theta)=\cfrac{hypotenuse}{adjacent}
\\\\\\
sec(u)=-2\iff sec(u)=-\cfrac{2}{1}\cfrac{\leftarrow hypotenuse}{\leftarrow adjacent}[/tex]
now... we have -2/1 but... which one is the negative? well
we know the angle is " π/2 < u < π", that simply means, the 2nd quadrant
well, in the 2nd quadrant, the "x" or adjacent side is negative, so the adjacent side is then -1
now, let us use the pythagorean theorem, to get the opposite side
[tex]\bf c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{(2)^2-(-1)^2}=b
\\\\\\
\pm\sqrt{3}=b[/tex]
now..which one though? well, recall, we're in the 2nd quadrant, in the 2nd quadrant "y" or the opposite side, is positive, so is √(3) then
now... let's plug in those values in the double-angle identities then
[tex]\bf \textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\\ \quad \\\\
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
\boxed{1-2sin^2(\theta)}\\
2cos^2(\theta)-1
\end{cases}
\\ \\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\
-----------------------------\\\\[/tex]
[tex]\bf sin(2u)=2\left( \cfrac{\sqrt{3}}{2} \right)\left( \cfrac{-1}{2}\right)
\\\\\\
cos(2u)=1-2\left( \cfrac{\sqrt{3}}{2} \right)^2
\\\\\\
tan(2u)=\cfrac{2\left( \frac{\sqrt{3}}{-1} \right)}{1-\left( \frac{\sqrt{3}}{-1} \right)^2}[/tex]
simplify away
