# Proving trigonometric equations:

• Apr 21st 2009, 02:26 PM
etheen
Can someone help me prove the following identities:

(1+cos^2x/1+tan^2x) tanx=cotx

e) secx-cosx=sinxtanx f)sinxcosx/sinx=1+tanx\tanx g)cosx/sinx-sinx/secx=cos^2xcotx

h)cotx-tanx/cotx+tanx=2cos^2x-1 i)3sin^2x+4cos2x=3+cos^2x j)9sec^2x-5tan^2x=5+4sec^2x

k) (2cos^2x-1)^2/sin^4x-cos^4x=2sin^2x-1 L)(sin^2x/1-cosx)-1=cosx

m)1+cos2x/sin2x=cotx

Okay, here are these ones for now... the rest are not the same so I want to have these down first:) Thank you so much for helping by the way! I wish I was a math wizz too!
• Apr 21st 2009, 10:11 PM
Trig Identities
Hello etheen
Quote:

Originally Posted by etheen
Can someone help me prove the following identities:

(1+cos^2x/1+tan^2x) tanx=cotx

This is not true. Try it when $x = \tfrac{\pi}{4}$ for example.

Quote:

e) secx-cosx=sinxtanx
$\sec x -\cos x = \frac{1}{\cos x}-\cos x= \frac{1-\cos^2x}{\cos x}$

Can you do it from here? (Use $\sin^2x + \cos^2x =1$)

Quote:

f)sinxcosx/sinx=1+tanx\tanx
Again, this is not true. The LHS is $\cos x$ and the RHS is $\cot x + 1$, and they are not identical.

Quote:

g)cosx/sinx-sinx/secx=cos^2xcotx
$\frac{\cos x}{\sin x}-\frac{\sin x}{\sec x}= \frac{\cos x}{\sin x}-\sin x\cos x$

$=\frac{\cos x - \sin^2 x \cos x}{\sin x}$

Factorise the numerator and then use $\sin^2x + \cos^2x =1$.

Hopefully, you've got the idea now, and can do some of the remaining questions yourself. Show us your attempts if you want more help.

• Apr 22nd 2009, 12:57 PM
etheen
okay for e) um....1-cos^2x =sin^2x and cosx=cottsint ...okay I'm stuck now:|

by the way what do you mean that f) is not true, grandad?
• Apr 22nd 2009, 01:14 PM
masters
Quote:

Originally Posted by etheen
okay for e) um....1-cos^2x =sin^2x and cosx=cottsint ...okay I'm stuck now:|

Hi etheen,

Here's the finish to (e). Work with the LHS

$\sec x -\cos x={\color{red}\sin x \tan x}$

$\frac{1}{\cos x}-\cos x=$

$\frac{1-\cos^2 x}{\cos x}=$

$\frac{\sin^2 x}{\cos x}=$

$\sin x \left(\frac{\sin x}{\cos x}\right)=$

${\color{red}\sin x \tan x}$
• Apr 22nd 2009, 01:42 PM
etheen
Hm, I see alittle!

can youplease exaplin exactly what explains when we do? 1/cosx-cosx?

you do LCD so we had to do 1 times cosx to get the denominator to be the same and once it was the same, we did cosx times cosx because we did that to the bototm right?
• Apr 22nd 2009, 09:58 PM
Trig Identities
Hello etheen
Quote:

Originally Posted by etheen
by the way what do you mean that f) is not true, grandad?

You are asking us to prove that
Quote:

f)sinxcosx/sinx=1+tanx\tanx
is true for all values of $x$ - and it simply isn't!

Are you sure you have written the question down correctly?

• Apr 23rd 2009, 11:47 AM
masters
Quote:

Originally Posted by masters
Hi etheen,

Here's the finish to (e). Work with the LHS

$\sec x -\cos x={\color{red}\sin x \tan x}$

$\frac{1}{\cos x}-\cos x=$

$\frac{1-\cos^2 x}{\cos x}=$

$\frac{\sin^2 x}{\cos x}=$

$\sin x \left(\frac{\sin x}{\cos x}\right)=$

${\color{red}\sin x \tan x}$

Quote:

Originally Posted by etheen
Hm, I see alittle!

can youplease exaplin exactly what explains when we do? 1/cosx-cosx?

you do LCD so we had to do 1 times cosx to get the denominator to be the same and once it was the same, we did cosx times cosx because we did that to the bototm right?

Does this help?

$\frac{1}{\cos x}-\cos x=\frac{1}{\cos x}-\frac{\cos^2 x}{\cos x}=\frac{1-\cos^2 x}{\cos x}$