Can someone explain how $\frac{\tan x}{\sec x}=\sin x$? What identities are used to get $\sin x$ from $\tan x \operatorname{/} \sec x$? I was looking at an example in my textbook and the problem went from $\tan x \operatorname{/} \sec x$ to $\sin x$. I don't understand how. 
 A: $$\tan x = \frac{\sin x}{\cos x} = \sin x \cdot \frac{1}{\cos x} = \sin x \sec x$$
Rearranging,
$$\sin x = \frac{\tan x}{\sec x}$$
A: Hints:

$\tan x=\frac{\sin x}{\cos x}$ and $\sec x=\frac{1}{\cos x}$

A: Here's an equivalent but more geometric answer. Let's say we have a right triangle with an angle $x$, the adjacent side with length $a$, the opposite side with length $b$ and the hypotenuse with length $c$.

Then $\tan x = \frac{b}{a}$, $\sec x = \frac{c}{a}$ and $\sin x = \frac{b}{c}$ (by definition of $\tan$, $\sec$, and $\sin$). That gives us
$$\frac{\tan x}{\sec x} = \frac{\frac{b}{a}}{\frac{c}{a}}=\frac{b}{a}\frac{a}{c}=\frac{b}{c}=\sin x$$
This method is also an easy way to derive the identity $\tan x = \frac{\sin x}{\cos x}$ that others are suggesting.
A: Just using the definitions:
$$
\frac{\tan x}{\sec x}=\frac{\frac{\sin x}{\cos x}}{\frac1{\cos x}}=\sin x.
$$
A: Rewrite $\tan x$ as $\dfrac{\sin x}{\cos x}$. This makes it easier to see: 
$$\tan x = \dfrac{\sin x}{\cos x} = \sin x \cdot \dfrac{1}{\cos x}$$ 
$$\implies = \sin x \sec x$$
$$\sin x = \frac{\tan x}{\sec x}$$
A: $\tan(x)=\frac{\sin(x)}{\cos(x)}$ and $\sec(x)=\frac{1}{\cos(x)}$
Therefore, $\frac{\tan(x)}{\sec(x)}=\frac{\sin(x)}{\cos(x)}(\frac{1}{\cos(x)})^{-1}$.
To divide fractions, you must multiply by the reciprocal, so the answer is $\sin(x)$.
A: $\tan(\pi/2) = \infty$ and $\sec(\pi/2)$ is undefined whereas $\sin(\pi/2) = 1$ so the given equation is incorrect.
