Problem: If derivative of $e^{ax} \cos{bx}$ with respect to $x$ is $re^{ax}\cos(bx + \tan^{-1} \frac {b} {a})$.
Then find $r$ when $a>0,b>0$
Solution: Differentiating $e^{ax} \cos{bx}$ w.r.t $x$ we get
$ ae^{ax} \cos{bx} -be^{ax} \sin{bx} $
But I don't know,how to convert it into $re^{ax}\cos(bx + \tan^{-1} \frac {b} {a})$
Hint: Use the identity
$$ \cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B). $$
to expand
$$ \cos(bx + \tan^{-1} \frac {b} {a}) $$
and then use the facts
$$ \cos(\arctan(t))=\frac{1}{\sqrt{1+t^2}},\quad \sin(\arctan(t))=\frac{t}{\sqrt{1+t^2}}. $$
Hint: Use trigonometric identities $\cos(\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta$ on $e^{ax}\left(a\cdot\cos(bx) - b\cdot\sin(bx)\right)$ (don't forget of limits for values of $a$ and $b$, while changing them into trigonometric functions).
