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

1.	If sec θ + tan θ = x, then tan θ =A. \(\frac{X^2+1}{X}\) B. \(\frac{X^2-1}{X}\) C. \(\frac{X^2+1}{2X}\) D. \(\frac{X^2-1}{2X}\)
Answer» Given: sec θ + tan θ = x ...…(i) To find: tan θ 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) Subtracting (ii) from (i), we get tanθ + tanθ = \(X-\frac{1}{X}\) 2tanθ = \(\frac{X^2-1}{X}\) tanθ = \(\frac{X^2-1}{2X}\)

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

Answer»

Given: sec θ + tan θ = x ...…(i)

To find: tan θ

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)

Subtracting (ii) from (i), we get

tanθ + tanθ = \(X-\frac{1}{X}\)

2tanθ = \(\frac{X^2-1}{X}\)

tanθ = \(\frac{X^2-1}{2X}\)