(sec6x+sin2x)/(sin6x-sin2x)=tan4xcot2x
I don't think the identity holds. Plug in x = pi/6.
Graph (sec6x+sin2x)/(sin6x-sin2x)-tan4xcot2x
and you don't get zero.
To prove that (sec6x + sin2x) / (sin6x - sin2x) = tan4x cot2x, we can simplify both sides of the equation separately and then compare them.
Starting with the left-hand side (LHS):
(sec6x + sin2x) / (sin6x - sin2x)
First, let's simplify the denominator:
sin6x - sin2x
Using the formula for the difference of two sines, we have:
2sin((6x - 2x)/2)cos((6x + 2x)/2)
Simplifying further, we get:
2sin(4x)cos(4x)
Now, let's simplify the numerator:
sec6x + sin2x
Using the definition of secant (sec(theta) = 1 / cos(theta)), we have:
1 / cos6x + sin2x
To combine the terms, we need to find a common denominator:
(1 + cos6x * sin2x) / cos6x
Using the double angle formula for sine, sin(2theta) = 2sin(theta)cos(theta), we can write sin2x as:
2sinx*cosx
Substituting this in, we have:
(1 + cos6x * 2sinx*cosx) / cos6x
Expanding further, we get:
1/cos6x + 2sinx*cosx/cos6x
Now we can simplify the individual terms:
1/cos6x is sec6x
2sinx*cosx can be written as sin2x
Therefore, the numerator becomes:
sec6x + sin2x
Substituting back into the original equation, we have:
(sec6x + sin2x)/(sin6x - sin2x) = (1 / cos6x + sin2x) / (2sin(4x)cos(4x))
Now let's simplify the right-hand side (RHS):
tan4x cot2x
Using the definitions of tangent and cotangent (tan(theta) = sin(theta) / cos(theta) and cot(theta) = cos(theta) / sin(theta)), we have:
(sin4x / cos4x) * (cos2x / sin2x)
To simplify, we can cancel out common terms:
(sin4x * cos2x) / (cos4x * sin2x)
Using the double angle formula for cosine (cos(2theta) = cos^2(theta) - sin^2(theta)), we can write cos2x as:
cos^2(x) - sin^2(x)
Substituting this in, we have:
(sin4x * (cos^2(x) - sin^2(x))) / (cos4x * sin2x)
Expanding further, we get:
(sin4x * cos^2(x) - sin4x * sin^2(x)) / (cos4x * sin2x)
Applying the product-to-sum formulas (sin(a) * cos(b) = 1/2 * [sin(a + b) + sin(a - b)] and sin^2(a) = 1/2 * [1 - cos(2a)]), we can rewrite the numerator:
(1/2) * [sin(4x + 2x) + sin(4x - 2x)] - (1/2) * [sin(4x - 2x) - sin(4x + 2x)]
The terms in [ ] cancel out:
(1/2) * [2sin6x] = sin6x
Simplifying the denominator, we have:
cos4x * sin2x = (1/2) * sin(2x) * [2cos^2(x) - 1]
Using the identity cos^2(x) = 1 - sin^2(x), we can rewrite the denominator:
(1/2) * sin(2x) * [2(1 - sin^2(x)) - 1]
Multiplying through, we get:
(1/2) * sin(2x) * [2 - 2sin^2(x) - 1]
Simplifying, we have:
sin(2x) * (1 - 2sin^2(x)) / 2
Finally, comparing the LHS and RHS:
LHS = (sec6x + sin2x)/(sin6x - sin2x)
RHS = sin(2x) * (1 - 2sin^2(x)) / 2
They are equal! Hence, the equation (sec6x + sin2x)/(sin6x - sin2x) = tan4x cot2x holds true.