Answer
The identity is verified.
$cos(\pi-θ)+sin(\frac{\pi}{2}-θ)=0$
Work Step by Step
$cos(u-v)=cos~u~cos~v+sin~u~sin~v$
$sin(u+v)=sin~u~cos~v+cos~u~sin~v$
$cos(\pi-θ)+sin(\frac{\pi}{2}-θ)=cos~\pi~cos~θ+sin~\pi~sin~θ+sin~\frac{\pi}{2}~cos~θ+cos~\frac{\pi}{2}~sin~θ=(-1)cos~θ+0~sin~θ+1~cos~θ+0~sin~θ=-cos~θ+cos~θ=0$
