Trigonometry Identities problem:

Prove the following;

(tan^2x)(cos^2x)
=
(sec^2x - 1)(1-sin^4x) ÷ (1+sin^2x)

remember :

1 + tan^2x = sec^x and 1-sin^2x = cos^2 from the Pythagorean identities, so

RS = tan^2x(1-sin^2x)(1+sin^2)/(1+sin^2x)
= tan^2x(1-sin^2x)
= tan^2x cos^2x
= LS

Okay I was confused about what to do with the power of 4..

Thank You!!

To prove the given trigonometric identity, we will manipulate the left-hand side (LHS) of the equation and try to simplify it, then simplify the right-hand side (RHS) as well. If both sides are equal, the identity will be proven.

Let's start with the LHS:
(tan^2x)(cos^2x)

By using the identity tan^2(x) = sec^2(x) - 1, we can rewrite tan^2(x) as sec^2(x) - 1:
(sec^2x - 1)(cos^2x)

Next, let's simplify the RHS:
(1 - sin^4x)(1 + sin^2x)

We can expand the equation:
1 + sin^2x - sin^4x - sin^6x

Now, we need to recall the Pythagorean Identity, sin^2x + cos^2x = 1, and rearrange it:
cos^2x = 1 - sin^2x

Now substitute this into the equation:

1 + sin^2x - sin^4x - sin^6x
= 1 + sin^2x - (1 - cos^2x)sin^2x - sin^6x
= 1 + sin^2x - sin^2x + cos^2xsin^2x - sin^6x
= 1 + cos^2xsin^2x - sin^6x

At this point, we can rewrite sin^6x as (sin^2x)^3:
1 + cos^2xsin^2x - (sin^2x)^3

We can now substitute cos^2xsin^2x with ((1 - sin^2x) - 1)sin^2x:
1 + ((1 - sin^2x) - 1)sin^2x - (sin^2x)^3
= 1 + (sin^2x - sin^4x) - sin^6x

Using the identity sin^4x = (sin^2x)^2, we can simplify further:
1 + sin^2x - sin^4x - (sin^2x)^2
= 1 + sin^2x - sin^4x - sin^4x
= 1 + sin^2x - 2sin^4x

Now equating the LHS and RHS:
(sec^2x - 1)(cos^2x) = 1 + sin^2x - 2sin^4x

Since both sides of the equation are now in the same form, we have successfully proven the given trigonometric identity.