Small angle approximation for $\frac{1+\sin\theta}{5+3\tan\theta-4\cos\theta}$

Question

Please could somebody explain how the expression involving $\theta$ that $$\frac{1+\sin\theta}{5+3\tan\theta-4\cos\theta}$$ approximates to for small values of $\theta$ is $1-2\theta+4\theta^2$?

Answer 1

Just compute the first terms of the Taylor series of $\dfrac{1+\sin\theta}{5+3\tan\theta-4\cos\theta}$, and you will get $1-2\theta+4\theta^2$.

Answer 2

Approximate the numerator and denominator up to $\theta^2$ terms and get $$\frac{1+\theta+o(\theta^2)}{1+3\theta+2\theta^2+o(\theta^2)}=(1+\theta)(1-3\theta+7\theta^2)+o(\theta^2)=1-2\theta+4\theta^2+o(\theta^2).$$