Given $f (x) = \tan(x)$.  

1. Use the fact that $\sin(x + y) = \sin(x) \cos(y) + \cos(x) \sin(y)$ and $\cos(x + y) = \cos(x) \cos(y) – \sin(x) \sin(y)$ to show that $\frac { f ( x + h ) - f ( x ) } { h } = ( \frac { 1 } { h } ) \frac { 1 } { \operatorname { cos } ^ { 2 } ( x ) \operatorname { cot } ( h ) - \operatorname { sin } ( x ) \operatorname { cos } ( x ) }$.

   Steps (a):

   $\frac{f(x+h)-f(x)}{h}=\frac{\text{tan}(x+h)-\text{tan}(x)}{h}=\left ( \frac{1}{h} \right )\left ( \frac{\text{sin}(x+h)}{\text{cos}(x+h)}-\frac{\text{sin}(x)}{\text{cos}(x)} \right )=$

2. At this point, can you evaluate $\lim\limits_ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$.

   Answer (b):

   No, you would have division by $0$.

3. Write the resulting equation when multiplying by $\frac { \tan ( h ) } { \tan ( h ) }$.

4. It is known that $\lim\limits _ { h \rightarrow 0 } \frac { \operatorname { tan } ( h ) } { h }=1$. Use this fact and your answer to part (c) to show that when $\lim\limits _ { h \rightarrow 0 } \frac { f ( x + h ) - f ( x ) } { h }$ for $f(x) = \tan(x)$ is applied, the result is $\sec^2(x)$.