Identities Crisis

**Freaky-Person** · January 1st 2007, 12:39 PM

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

**ThePerfectHacker** · January 1st 2007, 01:11 PM

Originally Posted by Freaky-Person

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

I presume,
$\frac{1}{1+\sin x}=\sec^2x - \frac{\tan x}{\cos x}$
Work one left hand side,
$\frac{1}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos^2 x}$
Split fraction,
$\frac{1}{\cos^2 x}-\frac{\sin x}{\cos ^2 x}=\sec^2 x-\frac{\tan x}{\cos x}$

**Soroban** · January 1st 2007, 05:03 PM

Hello, Freaky-Person!

Another approach . . .

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

Start with the right side . . .

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

Multiply top and bottom by $(1 + \sin x)\!:$

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

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