Question

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 1

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 = \(\frac{X^2+1}{2X}\)