Check out some similar questions!

hanzel · Feb 4, 2012, 03:54 AM

Please prove if these two are identity ty.

Aurora2000 · 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 · Mar 12, 2012, 06:49 PM

sec^4x-sec^2x=tan^4x+tan^2x

trigster · Apr 21, 2012, 06:28 AM

sin2x=2tanx/(1+tan^2x)

wendyjewel · Jun 9, 2012, 01:14 PM

4sin^2x+2cos^2x

jamesjonesmiami · Oct 16, 2012, 05:35 PM

I need to find the proof cot^2-1/csc^2=1-2sin^2

ebaines · 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