Answer
$u=-\dfrac{\sin (2x) \csc x}{2}$ and $v=\dfrac{\cos (2x) \csc x}{2}$
Work Step by Step
Rewrite the first equation as: $u=\dfrac{-v \sin 2x}{\cos 2x}$
Substitute the value of $u$ in the first equation to get the value of $v$.
$2v \times \dfrac{\sin^2 2x+\cos^2 2x}{\cos (2x)}=\sec x$
Therefore, the above equation yields:
$\dfrac{2v}{\cos x}=\sec x \implies v =\dfrac{\cos (2x) \csc x}{2}$
Substitute the value of $v$ in the first equation to get the value of $u$.
$u=\dfrac{-(\dfrac{\cos (2x) \csc x}{2}) \times \sin 2x}{\cos 2x}=-\dfrac{\sin (2x) \csc x}{2}$
So, the solution is $u=-\dfrac{\sin (2x) \csc x}{2}$ and $v=\dfrac{\cos (2x) \csc x}{2}$
