Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{\sin (x+a)}{\cos x}$

Let $f(x) = \frac{\sin (x+a)}{\cos x}$.
Using the quotient rule,$\frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Here,$u = \sin (x+a)$ and $v = \cos x$.
We know that $\frac{d}{dx} [\sin (x+a)] = \cos (x+a)$ and $\frac{d}{dx} [\cos x] = -\sin x$.
Substituting these into the quotient rule formula:
$f'(x) = \frac{\cos x \cdot \cos (x+a) - \sin (x+a) \cdot (-\sin x)}{\cos^2 x}$
$f'(x) = \frac{\cos x \cos (x+a) + \sin x \sin (x+a)}{\cos^2 x}$
Using the trigonometric identity $\cos A \cos B + \sin A \sin B = \cos (A - B)$,where $A = x+a$ and $B = x$:
$f'(x) = \frac{\cos ((x+a) - x)}{\cos^2 x}$
$f'(x) = \frac{\cos a}{\cos^2 x}$