1. ## Identities Crisis

1/1+Sin = Sec^2 - Tan/Cos

2. Originally Posted by Freaky-Person
1/1+Sin = Sec^2 - Tan/Cos
I presume,
$\displaystyle \frac{1}{1+\sin x}=\sec^2x - \frac{\tan x}{\cos x}$
Work one left hand side,
$\displaystyle \frac{1}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos^2 x}$
Split fraction,
$\displaystyle \frac{1}{\cos^2 x}-\frac{\sin x}{\cos ^2 x}=\sec^2 x-\frac{\tan x}{\cos x}$

3. Hello, Freaky-Person!

Another approach . . .

$\displaystyle \frac{1}{1 + \sin x} \:= \:\sec^2\!x - \frac{\tan x}{\cos x}$

$\displaystyle \sec^2\!x - \frac{\tan x}{\cos x} \;=\;\frac{1}{\cos^2\!x} - \frac{\frac{\sin x}{\cos x}}{\cos x}\;=\;\frac{1}{\cos^2x} - \frac{\sin x}{\cos^2\!x} \;=\;\frac{1 - \sin x}{\cos^2\!x}$
Multiply top and bottom by $\displaystyle (1 + \sin x)\!:$
. . $\displaystyle \frac{1 + \sin x}{1 + \sin x}\cdot\frac{1 - \sin x}{\cos^2\!x} \;=\; \frac{1 - \sin^2\!x}{(1 + \sin x)\cos^2\!x} \:=\:\frac{\cos^2\!x}{(1 + \sin x)\cos^2\!x} \:=\:\frac{1}{1 + \sin x}$