Answer
$\tan{\left(\frac{\pi}{2}+\theta\right)}=-\cot{\theta}$
Work Step by Step
Use a graphing utility to graph the given expression. (Refer to the graph below.)
Note that the graph is identical to the graph of $-\cot{\theta}$.
This means that $\tan{(\frac{\pi}{2}+\theta)}=-\cot{\theta}$.
RECALL:
$\tan{A}=\dfrac{\sin{A}}{\cos{A}}$
Use the definition above to obtain:
\begin{align*}
\tan{\left(\frac{\pi}{2}+\theta\right)}&=\dfrac{\sin{\left(\frac{\pi}{2}+\theta\right)}}{\cos{\left(\frac{\pi}{2}+\theta\right)}}
\end{align*}
RECALL:
(1) $\sin{(A+B)}=\sin{A}\cos{B}+\cos{A}\sin{B}$
(2) $\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$
Use the identities above twith $A=\dfrac{\pi}{2}$ and $B=\theta$ to obtain:
\begin{align*}
\dfrac{\sin{\left(\frac{\pi}{2}+\theta\right)}}{\cos{\left(\frac{\pi}{2}+\theta\right)}}&=\dfrac{\sin{\frac{\pi}{2}}\cos{\theta}+\cos{\frac{\pi}{2}}\sin{\theta}}{\cos{\frac{\pi}{2}}\cos{\theta}-\sin{\frac{\pi}{2}}\sin{\theta}}\\\\
&=\dfrac{1(\cos{\theta})+0(\sin{\theta})}{0(\cos{\theta})-1(\sin{\theta})}\\\\
&=\dfrac{\cos{\theta}+0}{0-\sin{\theta}}\\\\
&=\dfrac{\cos{\theta}}{-\sin{\theta}}\\\\
&=-\cot{\theta}
\end{align*}
Therefore,
$\tan{\left(\frac{\pi}{2}+\theta\right)}=-\cot{\theta}$
