# 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.

• Actually it isn't, when $x\equiv\frac\pi2\pmod\pi$. – Marc van Leeuwen Oct 3 '14 at 6:02
• the simplest thing to do is just to replace $\sec x$ with $\frac{1}{\cos x}$, and to note that $\frac{\tan x}{\sec x}=\sin x$ is equivalent to $\tan x = \sin x\sec x$ – John Joy Oct 8 '14 at 22:43

Hints:

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

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.

• +1. Elsewhere here, I'd been in a discussion of this type of manipulation. Your method is one I'll add to my bags of tricks. – JoeTaxpayer Oct 3 '14 at 16:25

Just using the definitions: $$\frac{\tan x}{\sec x}=\frac{\frac{\sin x}{\cos x}}{\frac1{\cos x}}=\sin x.$$

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

Rearranging,

$$\sin x = \frac{\tan x}{\sec x}$$

Rewrite $\tan x$ as $\dfrac{\sin x}{\cos x}$. This makes it easier to see: $$\tan x = \dfrac{\sin x}{\cos x} = \sin x \cdot \dfrac{1}{\cos x}$$ $$\implies = \sin x \sec x$$

$$\sin x = \frac{\tan x}{\sec x}$$

$\tan(x)=\frac{\sin(x)}{\cos(x)}$ 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 $\sin(x)$.

$\tan(\pi/2) = \infty$ and $\sec(\pi/2)$ is undefined whereas $\sin(\pi/2) = 1$ so the given equation is incorrect.

• True, but the equation holds outside the points of the form $\frac\pi2+n\pi$, $n\in\mathbb Z$, and these points are removable singularities of $\tan/\sec$. Also, the equation is true whenever both sides make sense. I would say that the given equation is correct under some small assumptions or in a suitable interpretation. – Joonas Ilmavirta Oct 3 '14 at 21:03