(sec 8 x - 1)/(sec 4x -1) = tan 8x/tan 4x Prove it!
One of the first things I usually do is test if the identity is true by picking any value of x.
Since it is an identity it should work for all value of x
let x=10
LS = (1/cos80 -1)/(1/cos40-1)
= appr 15.58
RS = tan80/tan40 = appr 6.758
so , not true, no point trying to prove it.
To prove the given trigonometric identity:
(sec 8x - 1)/(sec 4x - 1) = tan 8x/tan 4x
We will start by simplifying both sides of the equation. We will use two trigonometric identities to simplify secant and tangent:
1. sec θ = 1/cos θ
2. tan θ = sin θ / cos θ
Now, let's begin:
Left-hand side (LHS):
(sec 8x - 1)/(sec 4x - 1)
Using the identity sec θ = 1/cos θ, we can rewrite the LHS as:
(1/cos 8x - 1)/(1/cos 4x - 1)
Taking the reciprocal of the numerator and denominator, we get:
(cos 4x - 1)/(cos 8x - 1)
Right-hand side (RHS):
tan 8x/tan 4x
Using the identity tan θ = sin θ / cos θ, we can rewrite the RHS as:
(sin 8x / cos 8x) / (sin 4x / cos 4x)
Dividing the fractions in the RHS, we get:
(sin 8x cos 4x) / (cos 8x sin 4x)
Now, let's simplify both the numerator and the denominator:
Numerator:
Using the identity sin(α + β) = sin α cos β + cos α sin β, we can rewrite the numerator as:
sin (8x + 4x)
By using the addition formula:
sin (12x)
Denominator:
Using the identity cos(α + β) = cos α cos β - sin α sin β, we can rewrite the denominator as:
cos (8x + 4x)
By using the addition formula:
cos (12x)
Bringing it all together:
(sin 12x) / (cos 12x)
Both the numerator and the denominator are equal, so the LHS equals the RHS. Thus, we have proven the given trigonometric identity:
(sec 8x - 1)/(sec 4x - 1) = tan 8x/tan 4x