prove it if you can?

Question

-[tan^2x/sec^2x]+1=cos^2x

Answer 1

I'm basing my proof on the assumption that when you say "tan^2x" you are saying "tan squared x":

tan^2x = sin^2x/cos^2x and sec^2x = 1/cos^2x

- [(sin^2x/cos^2x) / (1/cos^2x)] + 1
which is the same thing as:

- [sin^2x/cos^2x times cos^2x/1] + 1

the cos^2x's can be eliminated by cross multiplication, leaving you with:

-(sin^2x) + 1 ... 1 - sin^2x

"1 - sin^2x" is equal to cos^2x (sin^2x + cos^2x = 1; cos^2x = 1 - sin^2x)

cos^2x = cos^2x

Answer 2

Both previous solutions direct you to the algebra and trig involve, but don't do a "proof". To prove the identity, you need to start on one side, use theorems and properties (ie algebra and trig identities) to arrive at the other side.

So, stealing from the second answer:

-[tan^2x/sec^2x]+1=-((sin²x/cos²x)/(1/cos²x))+1
-[tan^2x/sec^2x]+1= -((sin²x/cos²x)*cos²x)+1
-[tan^2x/sec^2x]+1=-sin²x+1
-[tan^2x/sec^2x]+1=1-sin²x
-[tan^2x/sec^2x]+1=cos^2x

Answer 3

Change the left side to sines and cosines, with tan x = sin x/cos x and sec x = 1/cos x:

-((sin²x/cos²x)/(1/cos²x))+1 = cos²x
-((sin²x/cos²x)*cos²x)+1 = cos²x
-sin²x+1=cos²x
1=cos²x+sin²x

Answer 4

ok. we have the original identity

sin^2x + cos^2x=1

subtract sin^2x

cos^2x= 1 - sin^2x

tan^2x/sec^2x =
(sin^2x/cos^2x)/(1/cos^2x)

since you are dividing by 1/cos^2x you multiply by cos^2x/1 (it's reciprocal)

sin^2x/cos^2x * cos^2x = sin^2x

this statemenet holds.

Answer 5

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

cos(x)^2 = 1 - sin(x)^2

-sin(x)^2 + 1 = 1 - sin(x)^2

so

(-(tan(x)^2)/(sec(x)^2)) + 1 = cos(x)^2