Answer
$u=1; v=-\tan x$
Work Step by Step
Rewrite the first equation as: $u=\dfrac{-v \cos x}{\sin x}$
Substitute the value of $u$ in the first equation to get the value of $v$.
$-v \times \dfrac{\cos^2 x+\sin^2 x}{\sin x}=\sec x$
Therefore, the above equation yields:
$v=-\sin x \sec x \implies v =\dfrac{-\sin x}{\cos x}=-\tan x$
Substitute the value of $v$ in the first equation to get the value of $u$.
$u=\dfrac{-\tan x \cos x}{\sin x}=1$
So, the solution is $u=1; v=-\tan x$
