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.
 A: $$\begin{align*}
\tan \theta \sin \theta + \cos \theta & \stackrel{\text{def.}}= \frac{\sin^2 \theta}{\cos \theta} + \cos \theta \\
& \stackrel{\text{Pythagoras}}= \frac{1-\cos^2 \theta}{\cos \theta} + \cos \theta \\
& = \frac1{\cos\theta} - \cos \theta + \cos \theta \\
& \stackrel{\text{def.}}= \sec\theta
\end{align*}$$
Where we use the definitions of $\tan \theta$ and $\sec\theta$ plus Pythagoras' theorem $\sin^2 \theta + \cos^2 \theta = 1$.
A: To Prove :$$\tan \theta\sin \theta + \cos \theta = \sec \theta$$
L.H.S :
$\tan \theta\sin \theta + \cos \theta $
$\implies \large\frac{\sin \theta}{\cos \theta}\times{\sin \theta} + \cos \theta $
$\implies \large \frac{\sin^2 \theta +\cos^2 \theta}{\cos \theta} $
$\implies \large \frac{1}{\cos \theta} $
$\implies \large {\sec \theta} $
:)
A: AS mentioned in a comment above
$$
\tan \theta =\dfrac{\sin \theta}{\cos \theta}.
$$
then your L.H.S becomes
$$
\dfrac{\sin \theta}{\cos \theta}.\sin \theta + \cos \theta
$$
from this can you add the terms together? and use $\sin^2 \theta +\cos^2 \theta = 1$
A: First note that
\[
\tan\theta =\frac{\sin\theta}{\cos\theta}
\]
\[
\sin^{2}\theta+\cos^{2}\theta=1
\]
\[
\frac{a}{b}\pm c=\frac{a\pm bc}{b}
\]
So then
\[
\tan\theta\sin\theta+\cos\theta= \frac{\sin^{2}\theta}{\cos\theta}+\cos\theta= \frac{\sin^{2}\theta+\cos^{2}\theta}{\cos\theta}=\frac{1}{\cos\theta}=\sec\theta
\]
