Proving Identities (Calculus)...?

Question

Okay so i have this one problem that i've been getting stuck on for a while now:

tan^2x - sin^2x = tan^2xsin^2x

Where: ^ means power

Notice: The left part is subtracting, the right part is multiplying

If anyone could help, I'd really appreciate it being solved from the left side, i've done my other problems from this side and just like consistency in my work. I'll try it also, and perhaps we could get the same answer.

Thank you for your time.

Answer 1

tan^2 - sin^2 = (sin^2/cos^2)(1-cos^2) = (sin^s/cos^2) sin^2
= tan^2x sin^2x

Answer 2

Ill dispose of the x, sin^2 means (sin(x))^2

tan^2 - sin^2 = (sin^2)/(cos^2) - sin^2
Getting a common denominator:
= (sin^2)/(cos^2) - ((sin^2)*(cos^2))/cos^2
= (sin^2 -sin^2 cos^2)/cos^2
Pulling out a sin^2 from the numerator:
=(sin^2*(1-cos^2))/cos^2
sin^2+cos^2 = 1, so 1-cos^2 = sin^2
(sin^2*(1-cos^2))/cos^2 = sin^4/cos^2
= (sin^2/cos^2)*sin^2
=tan^2*sin^2

In the future, don't be afraid to work from whatever side seems easiest; sometimes a solution is more evident in one direction than the other.

Answer 3

Is that a minus sign? i'm getting sin^4(x)sec^2(x) = Tan^2(x)sin^2(x). i think of that's what it incredibly is. if so basically understand your trig definitions. Tanx is basically sinx/cosx so tan ^2 x is sin^2x/cos^2 x. Secx is a million/cosx. So sec^2x is a million/cos^2x. Plug in values and you gets sin^4x and a million/cos^2x this is like I stated sec^2 x. Whabam!! wish I helped.

Answer 4

tan^2 x = (sin^2 x)/(cos^2 x) =>
tan^2x - sin^2x = (sin^2 x)[1/(cos^2 x) - 1] = (sin^2 x)[(1- cos^2 x)/(cos^2 x)] = (sin^2 x)[(sin^2 x)/(cos^2 x)] = (tan^2 x)(sin^2 x)

Answer 5

tan² x - sin² x
= sin² x / cos² x - sin² x
= (sin² x - sin² x.cos² x) / cos² x
= [sin² x (1 - cos² x)] / cos² x
= tan² x (1 - cos² x)
= tan² x.sin² x

Answer 6

sin^2x/cos^2x - sin^2x/1 = tan^2xsin^2x

sin^2x/cos^2x - sin^2xcos^2x/cos^2x = tan^2xsin^2x

(sin^2x -sin^2xcos^2x)/cos^2x

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

We use this identity to solve this:

sin^2x + cos^2x = 1

sin^2x = 1 - cos^2x

Now we continue:

sin^2x(sin^2x)/cos^2x

sin^2x(sin^2x/cos^2x)

sin^2x(tan^2x)

=tan^2xsin^2x

And now you're finished, ^_^

Answer 7

tan^2x - sin^2x = tan^2xsin^2x/COS^2x-sin^2x=sin^2x[1/COS^2x -1] sin^2x[TAN 2X[