## Trigonometry (11th Edition) Clone

$\tan{\left(\frac{\pi}{2}+\theta\right)}=-\cot{\theta}$
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, $\tan{\left(\frac{\pi}{2}+\theta\right)}=-\cot{\theta}$