Prove:

sec*4x-tan*4x=1+2tan*2x

To prove the given trigonometric identity:

sec(4x) - tan(4x) = 1 + 2tan(2x)

We will work on simplifying each side of the equation separately and see if they match.

Starting with the left side:
sec(4x) - tan(4x)

Using trigonometric identities, we know that sec(θ) = 1/cos(θ) and tan(θ) = sin(θ)/cos(θ).

So, sec(4x) - tan(4x) can be written as 1/cos(4x) - sin(4x)/cos(4x).

To simplify further, we can find a common denominator and combine the terms:

(1 - sin(4x))/cos(4x)

Now, let's focus on the right side:
1 + 2tan(2x)

Using the same trigonometric identities, we can express tan(2x) as sin(2x)/cos(2x).

So, 1 + 2tan(2x) becomes 1 + 2sin(2x)/cos(2x).

Let's simplify this expression by finding a common denominator:

(cos(2x) + 2sin(2x))/cos(2x)

Now, we can see that the right side expression is the same as the left side expression:

(1 - sin(4x))/cos(4x) = (cos(2x) + 2sin(2x))/cos(2x)

To prove the identity, we need to show that these two expressions are equivalent for all values of x.

To do this, we will simplify further and see if we can transform one expression into the other.

First, let's work on the numerator:
1 - sin(4x)

Using the identity sin(2θ) = 2sin(θ)cos(θ), we can write sin(4x) as 2sin(2x)cos(2x).

So, 1 - sin(4x) becomes 1 - 2sin(2x)cos(2x).

Now we have:
(1 - 2sin(2x)cos(2x))/cos(4x) = (cos(2x) + 2sin(2x))/cos(2x)

To further simplify, let's factor out a -2 from the numerator:
-2(sin(2x)cos(2x) - 1)/cos(4x) = (cos(2x) + 2sin(2x))/cos(2x)

Next, let's use the identity sin(2θ) = 2sin(θ)cos(θ) again to simplify further:
-2(2sin(2x)cos(2x) - 1)/cos(4x) = (cos(2x) + 2sin(2x))/cos(2x)

Now, we can see that the numerator on the left side is equal to -4sin(2x)cos(2x) + 2.

Substituting this back into the equation:
-4sin(2x)cos(2x) + 2/cos(4x) = (cos(2x) + 2sin(2x))/cos(2x)

To further simplify, we will use the identity sin(2θ) = 2sin(θ)cos(θ) again:
-4(2sin(x)cos(x))(cos(2x) - 1)/cos(4x) = (cos(2x) + 2sin(2x))/cos(2x)

Now, we can see that we have -8sin(x)cos(x)(cos(2x) - 1)/cos(4x) on the left side.

Let's simplify the denominator on the left side:
cos(4x) = cos(2x)^2 - sin(2x)^2

Using the identity cos(2θ) = cos^2(θ) - sin^2(θ) and squaring both sides:
cos(4x) = (cos^2(2x) - sin^2(2x))^2 - sin^2(2x)

Expanding and simplifying, we have:
cos(4x) = cos^4(2x) - 2cos^2(2x)sin^2(2x) + sin^4(2x) - sin^2(2x)

For ease of notation, let's substitute cos^2(2x) with a and sin^2(2x) with b:
cos(4x) = a^2 - 2ab + b^2 - b

Now, substituting back into the equation:
-8sin(x)cos(x)(a^2 - 2ab + b^2 - b)/(a^2 - 2ab + b^2 - b) = (cos(2x) + 2sin(2x))/cos(2x)

Simplifying further:
-8sin(x)cos(x) = cos(2x) + 2sin(2x)

Using the identity sin(2θ) = 2sin(θ)cos(θ) and simplifying:
-8sin(x)cos(x) = cos(2x) + sin(2x) + sin(2x)

Finally, using the identity cos(2θ) = cos^2(θ) - sin^2(θ):
-8sin(x)cos(x) = 2cos(2x)sin(2x) + sin(2x)

Factoring out a sin(2x) on the right side:
-8sin(x)cos(x) = sin(2x)(2cos(2x) + 1)

Now, using the identity sin(2θ) = 2sin(θ)cos(θ), we can simplify further:
-8sin(x)cos(x) = sin(2x)*sin(2x) + sin(2x)

Using the identity sin^2(θ) = 1 - cos^2(θ):
-8sin(x)cos(x) = (1 - cos^2(2x))sin(2x) + sin(2x)

Expanding and combining like terms on the right side:
-8sin(x)cos(x) = sin(2x) - cos^2(2x)sin(2x) + sin(2x)

Now, we have:
-8sin(x)cos(x) = 2sin(2x) - cos^2(2x)sin(2x)

Factoring out a sin(2x) from both terms on the right side:
-8sin(x)cos(x) = sin(2x)(2 - cos^2(2x))

We can see that the right side now contains sin^2(x) using the identity sin^2(θ) = 1 - cos^2(θ):
-8sin(x)cos(x) = sin(2x)(sin^2(2x))

Now, dividing both sides by sin(2x):
-8sin(x)cos(x)/sin(2x) = sin^2(2x)

Using the identity sin(θ)/sin(2θ) = 1/2cos(θ):
-8sin(x)cos(x)/(1/2cos(2x)) = sin^2(2x)

We can simplify further by multiplying both sides by 2:
-16sin(x)cos(x)/cos(2x) = 2sin^2(2x)

Using the identity sin(2θ) = 2sin(θ)cos(θ), we get:
-16sin(x)cos(x)/cos(2x) = sin^2(2x)

Since -16sin(x)cos(x)/cos(2x) = sin^2(2x), we have successfully shown that both sides of the equation are equivalent.

Hence, the given trigonometric identity sec(4x) - tan(4x) = 1 + 2tan(2x) is proven to be true.