Answer
$\frac{f\frac{\tan\theta-\tan t}{1+\tan\theta\tan t}+f\tan t}{h\tan\theta}$
Work Step by Step
Applying the difference formula for the tan function we get: $\frac{f(\tan(\theta-t)+f\tan t}{h\tan\theta}=\frac{f\frac{\tan\theta-\tan t}{1+\tan\theta\tan t}+f\tan t}{h\tan\theta}$