1. ## Verify this identity

$\frac{ \frac{1}{cos x}}{ \ \frac{1}{cos x} - \frac{sin x}{cos x}} = sec x(sec x+ tan x)$

I want to do this without turning the 1/cos x into sec x or sin x/cos x into tan x in the initial problem. Thanks!!!

2. Hello, Godzilla!

$\frac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x}} \:=\: \sec x(\sec x+ \tan x)$

I want to do this without turning the 1/cos x into sec x or sin x/cos x into tan x in the initial problem.
. . Why not? . . . It's the fastest way!
We have: . $\frac{\sec x}{\sec x - \tan x}$

$\text{Multiply by }\frac{\sec x + \tan x}{\sec x + \tan x}\!:\quad
\frac{\sec x}{\sec x - \tan x}\:\cdot\:\frac{\sec x + \tan x}{\sec x + \tan x}$

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

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Well, if you insist . . .

Multiply by $\frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\dfrac{1}{\cos x}\right)} {\cos x\left(\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x}\right)} \;\;=\;\;\frac{1}{1-\sin x}$

Multiply by $\frac{1+\sin x}{1+\sin x}\!:\quad \frac{1}{1-\sin x}\cdot\frac{1+\sin x}{1+\sin x} \;\;=\;\;\frac{1-\sin x}{1 - \sin^2\!x}\;\;=\;\;\frac{1+\sin x}{\cos^2\!x}$

. . . . . $=\;\;\frac{1}{\cos x}\left(\frac{1+\sin x}{\cos x}\right) \;\;=\;\;\frac{1}{\cos x}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) \;\;=\;\;\sec x(\sec x + \tan x)$