If tan θ + sec θ = x; what is the value of tanθ.secθ ?

\(\frac{1}{4}\)(x2 + 2x) \(\frac{1}{4}\)(x2 - 2x) \(\frac{1}{4}\)(x2 + \(\frac{1}{x^2}\)) \(\frac{1}{4}\)(x2 - \(\frac{1}{x^2}\))

\(\frac{1}{4}\)(x2 - \(\frac{1}{x^2}\))

tan θ + sec θ = x ⇒ tan θ - sec θ = \(\frac{1}{x}\) 4tanθ.secθ = (tanθ + secθ)2 - (tanθ - secθ)2 ⇒ 4tanθ.secθ = x2 - \(\frac{1}{x^2}\) ⇒ tanθ.secθ = \(\frac{1}{4}\)(x2 -  \(\frac{1}{x^2}\))