\(\cot\left(\theta+\dfrac{\pi}{2}\right)=-\tan\theta\\\\We\ know:\\\\\sin\left(x+\dfrac{\pi}{2}\right)=\cos x\\\\\cos\left(x+\dfrac{\pi}{2}\right)=-\sin x\\\\\tan x=\dfrac{\sin x}{\cos x}\\\\\cot x=\dfrac{\cos x}{\sin x}\\---------------\\\L=\cot\left(\theta+\dfrac{\pi}{2}\right)=\dfrac{\cos\left(\theta+\dfrac{\pi}{2}\right)}{\sin\left(\theta+\dfrac{\pi}{2}\right)}=\dfrac{-\sin\theta}{\cos\theta}=-\tan\theta=R\\\\\boxed{ }\)
