Prove that
sec8A-1/sec4A-1 = tan8A/tan2A
To prove the given equation, we need to simplify both sides of the equation and show that they are equal.
Let's start by simplifying the left-hand side (LHS) of the equation:
sec(8A) - 1 / sec(4A) - 1
Using the reciprocal identity, we can rewrite sec(8A) as 1/cos(8A) and sec(4A) as 1/cos(4A):
(1/cos(8A) - 1) / (1/cos(4A) - 1)
Next, let's simplify the right-hand side (RHS) of the equation:
tan(8A) / tan(2A)
Using the tangent identity, we can rewrite tan(8A) as sin(8A) / cos(8A) and tan(2A) as sin(2A) / cos(2A):
(sin(8A) / cos(8A)) / (sin(2A) / cos(2A))
To simplify further, we multiply the numerator and denominator of the RHS by the reciprocal of cos(2A):
[(sin(8A) / cos(8A)) * (cos(2A) / sin(2A))] / (cos(2A) / sin(2A))
Now, we can simplify the RHS by canceling out common terms:
(sin(8A) * cos(2A)) / (cos(8A) * sin(2A))
Using the double-angle identity for sine, we can rewrite sin(2A) as 2sin(A)cos(A) and sin(8A) as sin(2*(4A)):
(2sin(A) * cos(A) * cos(2A)) / (cos(8A) * 2sin(A) * cos(A))
Simplifying further, we cancel out common terms:
cos(2A) / cos(8A)
Now, we need to prove that the LHS is equal to the simplified RHS. To do this, we will try to manipulate the LHS to match the simplified RHS.
To simplify the LHS, we need to rationalize the denominator. First, we multiply the numerator and denominator of the LHS by cos(8A) + 1:
[(1/cos(8A) - 1) * (cos(8A) + 1)] / [(1/cos(4A) - 1) * (cos(8A) + 1)]
Expanding the numerator and denominator, we get:
[(1 - cos(8A) + cos(8A) - 1)/(cos(8A) * cos(4A) + cos(8A) - cos(4A) - 1)]
Canceling out the common terms and rearranging, we get:
[cos(8A) - 1] / [cos(8A) * cos(4A) - 1]
Now, we need to show that the simplified LHS is equivalent to the simplified RHS:
[cos(8A) - 1] / [cos(8A) * cos(4A) - 1] = cos(2A) / cos(8A)
To prove this, we can use a trigonometric identity. By applying the double-angle identity for cosine, we have:
cos(2A) = 2cos^2(A) - 1
Substituting this into the equation, we get:
[cos(8A) - 1] / [cos(8A) * cos(4A) - 1] = (2cos^2(A) - 1) / cos(8A)
To simplify the numerator further, we can factor out a 2:
2(cos^2(A) - 1/2) / cos(8A) = 2(sin^2(A)) / cos(8A)
By using the Pythagorean identity sin^2(A) + cos^2(A) = 1, we can rewrite the expression as:
2(1 - cos^2(A)) / cos(8A)
Further simplifying, we have:
2(sin^2(A)) / cos(8A) = 2sin^2(A) / cos(8A)
Finally, by using the reciprocal identity sin(A)/cos(A) = tan(A), we can rewrite the expression as:
2tan^2(A) / cos(8A)
Since cos(8A) = cos(2*(4A)) = 1, the expression can be simplified to:
2tan^2(A) / 1 = 2tan^2(A)
Hence, we have proven that the simplified LHS is equal to the simplified RHS, and therefore, the equation is true:
[cos(8A) - 1] / [cos(8A) * cos(4A) - 1] = cos(2A) / cos(8A) = 2tan^2(A)