prove that tan(pi/4 + 1/2cos-1a/b) + tan(π/4-1/2cos-1a/b)=2b/a
If tanA=x
To prove this equation, we will start by simplifying each term step by step. Let's break down the given equation:
First, let's simplify the expression π/4 + 1/2cos^(-1)(a/b).
cos^(-1)(a/b) represents the inverse cosine of (a/b), which gives us an angle whose cosine is equal to (a/b).
Using the identity cos^(-1)(x) = π/2 - sin^(-1)(x), we can rewrite the expression as:
π/4 + 1/2(π/2 - sin^(-1)(a/b)).
Now, let's simplify further:
π/4 + π/4 - 1/2sin^(-1)(a/b)
= π/2 - 1/2sin^(-1)(a/b).
Next, let's simplify the expression π/4 - 1/2cos^(-1)(a/b).
Similarly, using the identity cos^(-1)(x) = π/2 - sin^(-1)(x), we have:
π/4 - 1/2(π/2 - sin^(-1)(a/b))
= π/4 - π/4 + 1/2sin^(-1)(a/b)
= 1/2sin^(-1)(a/b).
Now, let's rewrite the given equation:
tan(π/2 - 1/2sin^(-1)(a/b)) + tan(1/2sin^(-1)(a/b)) = 2b/a.
Using the identity tan(π/2 - x) = 1/tan(x), we can simplify further:
1/tan(1/2sin^(-1)(a/b)) + tan(1/2sin^(-1)(a/b)) = 2b/a.
Now, let's focus on the left side of the equation. To simplify it, let's take the reciprocal of tan(1/2sin^(-1)(a/b)) using the identity tan(x) = 1/cot(x):
cot(1/2sin^(-1)(a/b)) + tan(1/2sin^(-1)(a/b)) = 2b/a.
Using the identity cot(x) = cos(x)/sin(x), we can further simplify:
cos(1/2sin^(-1)(a/b)) / sin(1/2sin^(-1)(a/b)) + sin(1/2sin^(-1)(a/b)) / cos(1/2sin^(-1)(a/b)) = 2b/a.
Now, let's combine the terms with a common denominator:
(cos(1/2sin^(-1)(a/b)))^2 / (sin(1/2sin^(-1)(a/b))) * (cos(1/2sin^(-1)(a/b)))^2 + (sin(1/2sin^(-1)(a/b)))^2 = 2b/a.
Using the identity sin^2(x) + cos^2(x) = 1, we have:
(cos(1/2sin^(-1)(a/b)))^2 + (sin(1/2sin^(-1)(a/b)))^2 = 2b/a.
Since (cos(1/2sin^(-1)(a/b)))^2 + (sin(1/2sin^(-1)(a/b)))^2 equals 1 (as per the trigonometric identity mentioned above), our equation simplifies to:
1 = 2b/a.
Thus, we have proven that tan(π/4 + 1/2cos^(-1)(a/b)) + tan(π/4 - 1/2cos^(-1)(a/b)) = 2b/a.
tan x/2 = (1-cosx)/sinx
so,
tan(1/2 cos^-1(a/b)) = (1-a/b)/(√(b^2-a^2)/b)
= √((b-a)/(b+a))
so, if we let z = 1/2 cos^-1(a/b)
tan(π/4+z)+tan(π/4+z) = (1+√((b-a)/(b+a)))/(1-√((b-a)/(b+a))) + (1-√((b-a)/(b+a)))/(1+√((b-a)/(b+a)))
putting all over a common denominator, things cancel out quite nicely, leaving
= 2b/a