1.

For a positive integer n, letfn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ)…(1+sec2nθ).Then

Answer»

For a positive integer n, let

fn(θ)=(tanθ2)(1+secθ)(1+sec2θ)(1+sec4θ)(1+sec2nθ).

Then





Discussion

No Comment Found