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...
Hello, aikenfan!
We're expected to know the basic identities and their variations.
. . $\displaystyle \sec^2\!x \:=\:\tan^2\!x + 1\quad\Rightarrow\quad\sec^2\!x - \tan^2\!x\:=\:1\quad\Rightarrow\quad \sec^2\!x - 1 \:=\:\tan^2\!x$
. . $\displaystyle \sin^2\!x + \cos^2\!x \:=\:1\quad\Rightarrow\quad 1-\sin^2\!x\:=\:\cos^2\!x\quad\Rightarrow\quad 1-\cos^2\!x\:=\:\sin^2\!x$
The left side is: .$\displaystyle \sec^7\!x\tan x - \sec^5\!x\tan x$$\displaystyle 16)\;\;\sec^6\!x(\sec x\tan x) - \sec^4\!x(\sec x\tan x) \;=\;\sec^5\!x\tan^3x$
Factor: .$\displaystyle \sec^5\!x(\underbrace{\sec^2\!x - 1}_{\text{This is }\tan^2\!x})\tan x$
And we have: .$\displaystyle \sec^5\!x\cdot\tan^2\!x\cdot\tan x \;=\;\sec^5\!x\tan^3x$
$\displaystyle 22)\;\;\frac{\sec x + \tan x}{\sec x - \tan x} \;=\;(\sec x +\tan x)^2$
Multiply the left side by $\displaystyle \frac{\sec x + \tan x}{\sec x + \tan x}$
. . $\displaystyle \frac{\sec x + \tan x}{\sec x - \tan x}\cdot{\color{blue}\frac{\sec x + \tan x}{\sec x + \tan x}} \;=\;\frac{(\sec x + \tan x)^2}{\underbrace{\sec^2\!x - \tan^2\!x}_{\text{This is 1}}} \;=\;(\sec x + \tan x)^2$
The left side is: .$\displaystyle \frac{\cos x\cdot\frac{\cos x}{\sin x}}{1-\sin x}-1 \;=\;\frac{\cos^2\!x}{\sin x(1-\sin x)}-1$$\displaystyle 25)\;\;\frac{\cos x\cot x}{1-\sin x} - 1 \;=\;\csc x$
Multiply the fraction by $\displaystyle \frac{1+\sin x}{1+\sin x}$
. . $\displaystyle \frac{\cos^2\!x}{\sin x(1-\sin x)}\cdot{\color{blue}\frac{1+\sin x}{1 + \sin x}} -1\;\;=\;\; \frac{\cos^2x(1 + \sin x)}{\sin x(\underbrace{1-\sin^2\!x}_{\text{This is }\cos^2\!x})} - 1$
. . $\displaystyle = \;\;\frac{\cos^2\!x(1+\sin x)}{\sin x\!\cdot\!\cos^2\!x}-1 \;\;=\;\;\frac{1 + \sin x}{\sin x} - 1\;\;=\;\;\frac{1}{\sin x} + \frac{\sin x}{\sin x} - 1$
. . $\displaystyle = \;\;\csc x + 1 - 1 \;\;=\;\;\csc x$