If y = sin(sin x), prove that d²y/dx² + tan x (dy/dx) + y cos 2x = 0

dy/dx = cos(sin x) × d/dx(sin x) = cos(sin x).cos x
d²y/dx² = -sin(sin x).cos²x - cos(sin x).sin x
Now, in the expression,
(-sin(sin x).cos²x - cos(sin x).sin x) + (tan x × cos(sin x)cos x) + sin(sin x)cos²(x) =
(-sin(sin x).cos²x - cos(sin x).sin x) + (sin x/cos x × cos(sin x)cos x) + sin(sin x)cos²(x) =
(-sin(sin x)cos²x + sin(sin x)cos²x - cos(sin x)sin x + cos(sin x)sin x) = 0
Hence proved.
dy/dx = cos(sin x) × d/dx(sin x) = cos(sin x).cos x
d²y/dx² = -sin(sin x).cos²x - cos(sin x).sin x
Coming to the expression,
(-sin(sin x).cos²x - cos(sin x).sin x) + (tan x × cos(sin x)cos x) + sin(sin x)cos²(x) =
(-sin(sin x).cos²x - cos(sin x).sin x) + (sin x/cos x × cos(sin x)cos x) + sin(sin x)cos²(x) =
(-sin(sin x)cos²x + sin(sin x)cos²x - cos(sin x)sin x + cos(sin x)sin x) = 0
Hence proved.