Simplify $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$ Please show how to simplify this type of expression without using a calculator. I'm new to trigonometry and I don't know how to simplify this expressions: $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$.
 A: Hint:
$\sec(x)=\displaystyle\frac{1}{\cos(x)}$, $\csc(x)=\displaystyle\frac{1}{\sin(x)}$, then use Pythagoras' Theorem. 
A: HINT:
$$\sec x=\frac{1}{\cos x}, \csc x= \frac{1}{\sin x}$$
Also, $$\sin^2 x+ \cos^2 x=1$$
A: $\left(\cos{\theta}\div\sec{\theta}\right)+\left(\sin{\theta}\div\csc{\theta}\right)=\cos{\theta}\div\frac{1}{\cos{\theta}}+\sin{\theta}\div\frac{1}{\sin{\theta}}=\cos^{2}{\theta}+\sin^{2}{\theta}=1$.
A: $$ \frac{\cos\theta}{\sec\theta} + \frac{\cos\theta}{\sec\theta} = \frac{\cos\theta}{\frac{1}{\cos\theta}} + \frac{\sin\theta}{\frac{1}{\sin\theta}} = \cos^{2}\theta + \sin^{2}\theta = 1$$
A: First, we need to make a common denominator of $\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}$.
Let's go!
$$\require{cancel}\begin{aligned}\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)-\sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\&=\frac{\cancel{\sin\theta}+\sin\theta\cos\theta\cancel{-\sin\theta}+\sin\theta\cos\theta}{1-\cos^2\theta}\\&=\frac{2\cancelto{1}{\sin\theta}\cos\theta}{\cancelto{\sin\theta}{\sin^2\theta}}\\&=\frac{2\cos\theta}{\sin\theta}\\&=2\cot\theta\end{aligned}$$
I hope this helps.
