# Trig Identities Question

• November 8th 2012, 05:51 PM
WhatthePatel
Trig Identities Question
sin2x-tan2x = -sin2xtan2x
Thats my question. And also i really suck at identities so if you can
maybe give me some tips on how to do better on them or how to make
them easier, would be helpful. Cause i usually get scared as soon as i see them.
Thank you.
• November 8th 2012, 06:02 PM
skeeter
Re: Trig Identities Question
$\sin^2{x} - \frac{\sin^2{x}}{\cos^2{x}} =$

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

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

$\frac{\sin^2{x}(\cos^2{x} - 1)}{\cos^2{x}}$

$\frac{\sin^2{x}(-\sin^2{x})}{\cos^2{x}}$

finish it ...
• November 8th 2012, 06:20 PM
WhatthePatel
Re: Trig Identities Question
Got it thanks