Trigonometry Identity: $\tan \theta\sin \theta + \cos \theta = \sec \theta$ Sorry if my question seems too simple. I cannot find a proof and my text book does not provide one either. I am supposed to prove:
$$\tan \theta \times \sin \theta + \cos \theta = \sec \theta$$
I know that $\sec = \frac{1}{\cos\theta}$. But I do not know how to prove that $\tan \theta \times \sin \theta + \cos \theta = \frac{1}{\cos \theta}$. 
I appreciate if someone point me to the right direction.

$$\begin{align*}
\tan \theta \sin \theta + \cos \theta & \stackrel{\text{def.}}= \frac{\sin^2 \theta}{\cos \theta} + \cos \theta \\
& \stackrel{\text{Pythagoras}}= \frac{1-\cos^2 \theta}{\cos \theta} + \cos \theta \\
& = \frac1{\cos\theta} - \cos \theta + \cos \theta \\
& \stackrel{\text{def.}}= \sec\theta
\end{align*}$$
Where we use the definitions of $\tan \theta$ and $\sec\theta$ plus Pythagoras' theorem $\sin^2 \theta + \cos^2 \theta = 1$.
