I have the following identity:
$$ \frac{\tan (t + h) - \tan(t)}{h} = \left( \frac{\tan (h)}{h} \right)\left( \frac{\sec^2(t)}{1 - \tan (t)\tan (h)} \right)$$
Having tried various approaches, which are far too varied and numerous to list all here, the only one that seems to have the most promise is this one (however it still falls quite short); using the right hand side:
$$\begin{align}\text{LHS} &= \left( \frac{\tan (h)}{h} \right)\left( \frac{\sec^2(t)}{1 - \tan (t)\tan (h)} \right)\\
&=\frac{\dfrac{\sin (h)}{\cos(h)}}{h} \cdot \dfrac{\dfrac{1}{\cos^2(t)}}{\dfrac{\cos(t)\cos(h)-\sin(t)\sin(h)}{\cos(t)\cos(h)}}\\
&= \dfrac{1}{h} \cdot \dfrac{\dfrac{\sin(h)}{\cos(h)\cos^2(t)}}{\dfrac{\cos(t)\cos(h)-\sin(t)\sin(h)}{\cos(t)\cos(h)}}\\
&= \frac{1}{h} \cdot \frac{\sin (h)}{\cos(h)\cos ^2(t)}\cdot \frac{\cos (t)\cos(h)}{\cos(t)\cos(h)-\sin(t)\sin(h)}\\
&= \frac{\sin(h)}{h\cdot\cos(t)\cdot\cos(t+h)}\end{align}$$
Can anyone throw me a bone here? I'm stumped. Thanks.
Answer
Apply the rule
$$ \tan(t+h) = \frac{\tan(t) + \tan(h)}{1-\tan(t)\tan(h)} $$
to the LHS and it's straightforward from there.
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