hanzel Posts: 1, Reputation: 1 New Member #1 Feb 4, 2012, 03:54 AM
Proving trigonometric identities solver?
Please prove if these two are identity ty.

 Aurora2000 Posts: 111, Reputation: 16 Junior Member #2 Feb 4, 2012, 07:53 AM
$\frac{\cot x}{\csc x+1}=\sec x-\tan x$

$\frac{\cot x}{\csc x+1}=\frac{1}{\tan x(\csc x+1) }=\frac{1}{\tan x\csc x+\tan x}$
Now $\tan x\csc x=\tan x\sin^{-1} x=\cos^{-1} x=\sec x$, thus
$\frac{1}{\tan x\csc x+\tan x}=\frac{1}{\sec x+\tan x}$
Now you have to prove
$\frac{1}{\sec x+\tan x}=\sec x-\tan x$.
Multiplying both sides by $\sec x+\tan x$ this becomes
$1=\sec^2x-\tan^2x =\frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}=\frac{1-\sin^2x}{\cos^2x}$
and using $1=\sin^2x+\cos^2x$ you conclude.

 Jmiller23 Posts: 1, Reputation: 1 New Member #3 Mar 12, 2012, 06:49 PM
sec^4x-sec^2x=tan^4x+tan^2x
 trigster Posts: 1, Reputation: 1 New Member #4 Apr 21, 2012, 06:28 AM
sin2x=2tanx/(1+tan^2x)
 wendyjewel Posts: 1, Reputation: 1 New Member #5 Jun 9, 2012, 01:14 PM
4sin^2x+2cos^2x
 jamesjonesmiami Posts: 1, Reputation: 1 New Member #6 Oct 16, 2012, 05:35 PM
I need to find the proof cot^2-1/csc^2=1-2sin^2
 ebaines Posts: 12,132, Reputation: 1307 Expert #7 Oct 17, 2012, 06:05 AM
Originally Posted by jamesjonesmiami
i need to find the proof cot^2-1/csc^2=1-2sin^2
You have mistated the identy: it should be (cot^2-1)/csc^2 = 1-2sin^2

As always, it's best to replace the cot and csc functions with their sin and cos equivalents. Then it comes right out if you apply one of the standard iidentities that you should know be heart: cos^2 + sin^2 = 1

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