SOLUTION: Ok I'm having trouble proving these harder identities. The homework says prove that each of the following identities is true (2sec^2)x-(2sec^2)x(sin^2)x-(sin^2)x-(cos^2)x=1 It's

Algebra ->  Trigonometry-basics -> SOLUTION: Ok I'm having trouble proving these harder identities. The homework says prove that each of the following identities is true (2sec^2)x-(2sec^2)x(sin^2)x-(sin^2)x-(cos^2)x=1 It's      Log On


   



Question 368957: Ok I'm having trouble proving these harder identities.
The homework says prove that each of the following identities is true
(2sec^2)x-(2sec^2)x(sin^2)x-(sin^2)x-(cos^2)x=1
It's w/o the parenthesis in the homework, I just didn't know how to seperate the squares for the x's.
I tried multiplying by (2cos^2)x to get rid of the sec's but I'm getting nowhere with it. Here's what I did even though I know it's wrong:
*-1 then (2cos^2)x
=-(2cos^4)x+(4cos^2)x(sin^2)x=0
/(2cos^2)x
=(-cos^2)x+(2cos^2)x(sin^2)x=0
Then I don't see anywhere to go!

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(2sec^2)x-(2sec^2)x(sin^2)x-(sin^2)x-(cos^2)x=1
-------------------------------------------------
2sec^2 - 2sec^2*sin^2 -[sin^2+cos^2] = 1
----
2sec^2[1-sin^2] - 1 = 1
----
2sec^2[cos^2] = 2
---
2(1) = 2
2 = 2
==========
Cheers,
Stan H.