(a) Differentiate sec⁻¹(1/√(1-x²)) w.r.t. sin⁻¹(2x√(1-x²))

Let u = sec⁻¹(1/√(1-x²)) and v = sin⁻¹(2x√(1-x²))

First, simplify u: u = sec⁻¹(1/√(1-x²)) = cos⁻¹(√(1-x²))

Let x = sinθ, then u = cos⁻¹(√(1-sin²θ)) = cos⁻¹(cosθ) = θ = sin⁻¹(x)

Now, simplify v: v = sin⁻¹(2x√(1-x²))

Let x = sinθ, then v = sin⁻¹(2sinθ√(1-sin²θ)) = sin⁻¹(2sinθcosθ) = sin⁻¹(sin2θ) = 2θ = 2sin⁻¹(x)

Now, differentiate u and v with respect to x:

du/dx = d(sin⁻¹(x))/dx = 1/√(1-x²)

dv/dx = d(2sin⁻¹(x))/dx = 2/√(1-x²)

We need to find du/dv = (du/dx) / (dv/dx) = (1/√(1-x²)) / (2/√(1-x²)) = 1/2

(b) If y = tan x + sec x, then prove that d²y/dx² = cos x / (1 - sin x)²

Given y = tan x + sec x

First, find dy/dx: dy/dx = d(tan x + sec x)/dx = sec²x + sec x tan x = sec x (sec x + tan x)

Now, find d²y/dx²: d²y/dx² = d(sec x (sec x + tan x))/dx

Using the product rule: d²y/dx² = sec x tan x (sec x + tan x) + sec x (sec x tan x + sec²x)

d²y/dx² = sec²x tan x + sec x tan²x + sec²x tan x + sec³x = 2sec²x tan x + sec x tan²x + sec³x

d²y/dx² = sec x (2sec x tan x + tan²x + sec²x) = sec x (2sec x tan x + (sec²x - 1) + sec²x) = sec x (2sec x tan x + 2sec²x - 1)

d²y/dx² = (1/cos x) * [2(1/cos x)(sin x/cos x) + 2(1/cos²x) - 1] = (1/cos x) * [2sin x/cos²x + 2/cos²x - 1]

d²y/dx² = (1/cos x) * [(2sin x + 2 - cos²x) / cos²x] = (2sin x + 2 - (1 - sin²x)) / cos³x = (2sin x + 1 + sin²x) / cos³x

d²y/dx² = (sin²x + 2sin x + 1) / cos³x = (sin x + 1)² / cos³x = (sin x + 1)² / [cos²x * cos x] = (sin x + 1)² / [(1 - sin²x) * cos x]

d²y/dx² = (sin x + 1)² / [(1 - sin x)(1 + sin x) * cos x] = (sin x + 1) / [(1 - sin x) * cos x] = (1 + sin x) / [(1 - sin x) * cos x]

d²y/dx² = (1 + sin x) / [(1 - sin x) * cos x] * [(1 - sin x) / (1 - sin x)] = (1 - sin²x) / [(1 - sin x)² * cos x] = cos²x / [(1 - sin x)² * cos x]

d²y/dx² = cos x / (1 - sin x)²