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