Show that: tan(1/2 sin⁻¹(3/4)) = 4 - √73

Let θ = sin⁻¹(3/4)
We know that sin x = tan x √(1 - x²)
Therefore, θ = sin⁻¹(3/4) = tan⁻¹(3/4)√(1 - (3/4)²)
θ = tan⁻¹(3/4)√(7/16) = tan⁻¹(3√7/4)
θ = tan⁻¹(3√7)
tan θ = 3√7
We know that tan x = (2 tan(x/2)) / (1 - tan²(x/2))
tan θ = 2 tan(θ/2) / (1 - tan²(θ/2)) = 3√7
3 tan²(θ/2) = 2√7 tan(θ/2) + 3√7
3 tan²(θ/2) - 2√7 tan(θ/2) - 3√7 = 0
tan(θ/2) = (2√7 ± √(28 + 36 * 3√7)) / 6
tan(θ/2) = (2√7 ± √(28 + 108√7)) / 6
Since sin(θ/2) is acute, tan(θ/2) which is the R.H.S of the expression.
Since we started with substitution θ = sin⁻¹(3/4)
L.H.S = tan(1/2 sin⁻¹(3/4)) = 4 - √73
tan(θ/2) = 4 - √73, which is the L.H.S. of the expression.
Therefore, L.H.S = R.H.S.