## Algebra 2 (1st Edition)

Plugging in $t=0$ into the simplified formula $\frac{f\frac{\tan\theta-\tan t}{1+\tan\theta\tan t}+f\tan t}{h\tan\theta}$ we get: $\frac{f\frac{\tan\theta-\tan (0)}{1+\tan\theta\tan (0)}+f\tan (0)}{h\tan\theta}=\frac{f\frac{\tan\theta-0}{1+\tan\theta(0)}+f(0)}{h\tan\theta}=\frac{f\frac{\tan\theta}{1}}{h\tan\theta}=\frac{f}{h}$ Hence we proved what we had to.