<s>[INST] Evaluating $ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $ 
I need help with this limit:
  $$ \lim_{a \to 0} \frac{\tan(x+a)-\tan x}{a} $$

I've tried using the $$ \tan(a-b) = \frac{\tan a-\tan b}{1+\tan a\tan b} $$ with $(x+a)$ as $a$ and $(x)$ as $b$ but having difficulty getting beyond that.
 [/INST] $$\lim_{a \to 0} \frac{\tan(x+a)- \tan x}{a}= \lim_{a \to 0} \frac{\frac{\sin(x+a)}{\cos(x+a)}- \frac{\sin x}{\cos x}}{a}\\
=\lim_{a \to 0} \frac{\cos x \sin(x+a)-\sin x \cos (x+a)}{a(\cos x \cos (x+a))}\\
= \lim_{a \to 0} \frac{\cos x (\sin x \cos a + \cos x \sin a) - \sin x (\cos x \cos a - \sin x \sin a)}{a\cos x \cos (x+a)}\\
= \lim_{a \to 0} \frac{\sin a (\cos^2 x+ \sin^2 x)}{a \cos x \cos (x+a)}= \lim_{a \to 0} \frac{\sin a}{a} \frac {1}{\cos x \cos (x+a)}= \frac {1}{\cos^2 x}= \sec^2 x.$$
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