If sec θ + tan θ = x, then sec θ =A. \(\frac{X^2+1}{X}\) B. \(\f

1.	If sec θ + tan θ = x, then sec θ =A. \(\frac{X^2+1}{X}\) B. \(\frac{X^2+1}{2X}\) C. \(\frac{X^2-1}{2X}\) D. \(\frac{X^2-1}{X}\)
Answer» Given: secθ + tanθ = x …(i) To find: secθ We know that 1 + tan²θ = sec²θ ⇒ sec²θ – tan²θ = 1 ∵ a² – b² = (a – b) (a + b) ∴ sec²θ – tan²θ = (secθ – tanθ) (secθ + tanθ) = 1 ⇒ From (i), we have ⇒ (secθ – tanθ) x = 1 ⇒ secθ – tanθ = \(\frac{1}{x}\) ...…(ii) Adding (i) and (ii), we get secθ + secθ = \(X+\frac{1}{X}\) ⇒ secθ = \(\frac{X^2+1}{2X}\) ⇒ secθ = \(\frac{X^2+1}{2X}\)

If sec θ + tan θ = x, then sec θ =A. \(\frac{X^2+1}{X}\) B. \(\frac{X^2+1}{2X}\) C. \(\frac{X^2-1}{2X}\) D. \(\frac{X^2-1}{X}\)

Answer»

Given: secθ + tanθ = x …(i)

To find: secθ

We know that 1 + tan²θ = sec²θ

⇒ sec²θ – tan²θ = 1

∵ a² – b² = (a – b) (a + b)

∴ sec²θ – tan²θ = (secθ – tanθ) (secθ + tanθ) = 1

⇒ From (i), we have

⇒ (secθ – tanθ) x = 1

⇒ secθ – tanθ = \(\frac{1}{x}\) ...…(ii)

Adding (i) and (ii), we get

secθ + secθ = \(X+\frac{1}{X}\)

⇒ secθ = \(\frac{X^2+1}{2X}\)