Assume, $f(x) = \frac{{\sin (x + a)}}{{\cos x}}$.

Upon differentiating with respect to $x$ and applying quotient rule we get,

${f^\prime }(x) = \frac{{\cos x\frac{d}{{dx}}[\sin (x + a)] – \sin (x + a)\frac{d}{{dx}}\cos x}}{{{{\cos }^2}x}}$

${f^\prime }(x) = \frac{{\cos x\frac{d}{{dx}}[\sin (x + a)] – \sin (x + a)( – \sin x)}}{{{{\cos }^2}x}}$          …… (1)

Also, assume $g(x) = \sin (x + a)$ then,

$g(x + h) = \sin (x + h + a)$

By the first principle we have,

${g^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{g(x + h) – g(x)}}{h}$

$g'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}[\sin (x + h + a) – \sin (x + a)]$

$g'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{x + h + a + x + a}}{2}} \right)\sin \left( {\frac{{x + h + a – x – a}}{2}} \right)} \right]$

$g'(x) = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}\left[ {2\cos \left( {\frac{{2x + 2a + h}}{2}} \right)\sin \left( {\frac{h}{2}} \right)} \right]$

$g'(x) = \mathop {\lim }\limits_{h \to 0} \left[ {\cos \left( {\frac{{2x + 2a + h}}{2}} \right)\left\{ {\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right\}} \right]$

Now, applying the limits as $h \to 0 \Rightarrow \frac{h}{2} \to 0$ then,

$g'(x) = \mathop {\lim }\limits_{h \to 0} \cos \left( {\frac{{2x + 2a + h}}{2}} \right) \cdot \mathop {\lim }\limits_{\frac{f}{2} \to 0} \left\{ {\frac{{\sin \left( {\frac{h}{2}} \right)}}{{\left( {\frac{h}{2}} \right)}}} \right\}$

Put $\mathop {\lim }\limits_{h \to 0} \frac{{\sin h}}{h} = 1$ then,

$g'(x) = \left( {\cos \frac{{2x + 2a}}{2}} \right) \times 1$

$g'(x) = \cos (x + a)$                                          …… (2)

Substituting equation (2) in equation (1) to get,

${f^\prime }(x) = \frac{{\cos x \cdot \cos (x + a) + \sin x\sin (x + a)}}{{{{\cos }^2}x}}$

$f'(x) = \frac{{\cos (x + a – x)}}{{{{\cos }^2}x}}$

Therefore, $f'(x) = \frac{{\cos a}}{{{{\cos }^2}x}}$.