Help solve this trig identity?

Question

sec^2x*sin^2x-sec^2x=-1
I got a test and can't solve this one on my review...
Thanks!

Answer 1

just remember sec^2 2x=1/cos^2 2x. Multiply both sides of the equation by cos^2 2x and see what you get.

Also remember sin^2 A + cos^2 A=1

Answer 2

sec^2x*sin^2x-sec^2x=-1
sec^2 x * sin^2 x - sec^2 x = -1
sec^2 x (sin^2 x - 1) = -1
sin^2 x - 1 = -1/sec^2 x
sin^2 x - 1 = -cos^2 x (because 1/sec = cos)
sin^2 x + cos^2 x = 1
1 = 1

Answer 3

before solving we need to remember some of the simple trig identidies.
we know that secX = 1/ cosX
we know that sinX/cosX = tanX
and we know that tan^2X + 1 = Sec ^2X.

okay.

so here is the equation:
(sec^2X)(sin^2X)-sec^2X= -1

we are going to change the first "sec^2X" into "1/cos^2X"
so now we have:

(___1____)(sin^2X)-sec^2X= -1
cos^2X

if you multiply (1/cos^2x) and (sin ^2X) you will get:

(sin^2X/cos^2X) - sec ^2X = -1

we can change "sin^2X/cos^2X" into "tan^2X" and now we have:

tan^2X - sec^2X = -1

if you add one to both sides you have:

tan^2X - sec^2X +1 = 0

now add sec^2X to both sides and you get:

tan^2X +1 = sec^2X

now we know (as stated in the first sentence of this answer) that "tan^2X +1 = Sec^2X"

therefore you have:

sec^2X = sec^2X

done.