Question

If `(tan^3θ - 1)/(tan θ - 1) = A sec^2 θ + B tan θ`, then A + B = ______.

Answer 1

If `(tan^3θ - 1)/(tan θ - 1) = A sec^2 θ + B tan θ`, then A + B = 2.

Explanation:

Let t = tan θ. Then

`(t^3 - 1)/(t - 1) = t^2 + t + 1`   ...(Since a³ – b³ = (a – b)(a² + ab + b²))

Now tan² θ = sec² θ – 1   ...(∵ sec² θ = 1 + tan² θ)

So t² + t + 1 = (sec² θ – 1) + tan θ + 1

= sec² θ + tan θ

Thus A = 1 and B = 1, so A + B = 2.