# Math Help - Trig Functions

1. ## Trig Functions

I had to simplify all of these (so that the RS and LS are equal)...I was wondering if someone would please check my worK? The original part of the problem is next to the number...all work under that is my own...

2. 1, 4, 7, 10, 13, 19 are all correct. (as far as content is concerned)

I'm not sure what is going on with number 25, do you know how to do it?

I can't read #16, and 22, if somone else could read them that would be awsome.

3. On 16, 25, and 22, I do not know how to do them, I was experimenting with 22, but I don't know the correct way to do it...I very much appreciate you taking the time to look over my work....

4. Can anyone please help me with any of those? I attempted at them, but I can't seem to get it right

5. Hello, aikenfan!

We're expected to know the basic identities and their variations.

. . $\sec^2\!x \:=\:\tan^2\!x + 1\quad\Rightarrow\quad\sec^2\!x - \tan^2\!x\:=\:1\quad\Rightarrow\quad \sec^2\!x - 1 \:=\:\tan^2\!x$

. . $\sin^2\!x + \cos^2\!x \:=\:1\quad\Rightarrow\quad 1-\sin^2\!x\:=\:\cos^2\!x\quad\Rightarrow\quad 1-\cos^2\!x\:=\:\sin^2\!x$

$16)\;\;\sec^6\!x(\sec x\tan x) - \sec^4\!x(\sec x\tan x) \;=\;\sec^5\!x\tan^3x$
The left side is: . $\sec^7\!x\tan x - \sec^5\!x\tan x$

Factor: . $\sec^5\!x(\underbrace{\sec^2\!x - 1}_{\text{This is }\tan^2\!x})\tan x$

And we have: . $\sec^5\!x\cdot\tan^2\!x\cdot\tan x \;=\;\sec^5\!x\tan^3x$

$22)\;\;\frac{\sec x + \tan x}{\sec x - \tan x} \;=\;(\sec x +\tan x)^2$

Multiply the left side by $\frac{\sec x + \tan x}{\sec x + \tan x}$

. . $\frac{\sec x + \tan x}{\sec x - \tan x}\cdot{\color{blue}\frac{\sec x + \tan x}{\sec x + \tan x}} \;=\;\frac{(\sec x + \tan x)^2}{\underbrace{\sec^2\!x - \tan^2\!x}_{\text{This is 1}}} \;=\;(\sec x + \tan x)^2$

$25)\;\;\frac{\cos x\cot x}{1-\sin x} - 1 \;=\;\csc x$
The left side is: . $\frac{\cos x\cdot\frac{\cos x}{\sin x}}{1-\sin x}-1 \;=\;\frac{\cos^2\!x}{\sin x(1-\sin x)}-1$

Multiply the fraction by $\frac{1+\sin x}{1+\sin x}$

. . $\frac{\cos^2\!x}{\sin x(1-\sin x)}\cdot{\color{blue}\frac{1+\sin x}{1 + \sin x}} -1\;\;=\;\; \frac{\cos^2x(1 + \sin x)}{\sin x(\underbrace{1-\sin^2\!x}_{\text{This is }\cos^2\!x})} - 1$

. . $= \;\;\frac{\cos^2\!x(1+\sin x)}{\sin x\!\cdot\!\cos^2\!x}-1 \;\;=\;\;\frac{1 + \sin x}{\sin x} - 1\;\;=\;\;\frac{1}{\sin x} + \frac{\sin x}{\sin x} - 1$

. . $= \;\;\csc x + 1 - 1 \;\;=\;\;\csc x$