## Precalculus (10th Edition)

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We know that $\sin{(\theta)}=\cos{(90^{o}-\theta)}$, and $\cot{(\theta)}=\frac{\sin{(\theta)}}{\cos{(\theta)}}$, hence $\cot{(40^{o})}-\frac{\sin{(50^{o})}}{\sin{(40^{o})}}=\cot{(40^{o})}-\frac{\cos{(40^{o})}}{\sin{(40^{o})}}=\cot{(40^{o})}-\cot{(40^{o})}=0.$