Verified Answer :

dalendrk

\(\tan\left(x+\dfrac{\pi}{2}\right)=-\cot(x)\\\\L=\dfrac{\sin\left(x+\frac{\pi}{2}\right)}{\cos\left(x+\frac{\pi}{2}\right)}=\dfrac{\sin(x)\cos\frac{\pi}{2}+\sin\frac{\pi}{2}\cos(x)}{\cos(x)\cos\frac{\pi}{2}-\sin(x)\sin\frac{\pi}{2}}\\\\=\dfrac{\sin(x)\cdot0+1\cdot\cos(x)}{\cos(x)\cdot0-\sin(x)\cdot1}=\dfrac{\cos(x)}{-\sin(x)}=-\cot(x)=R\\\\\\Used:\\\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)\\\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)\\\cot(x)=\dfrac{\cos(x)}{\sin(x)}\\\sin\frac{\pi}{2}=1\\\cos\frac{\pi}{2}=0\)

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