If \(\frac{{sec\theta + tan\theta }}{{sec\theta - tan\theta }} = 5\)�

1.	If \(\frac{{sec\theta + tan\theta }}{{sec\theta - tan\theta }} = 5\) and θ is an acute angle, then the value of \(\frac{{3{{\cos }^2}\theta + 1}}{{3{{\cos }^2}\theta - 1}}\) is: 1. 42. 33. 14. 2
Answer» Correct Answer - Option 1 : 4 Given: \(\frac{{secθ + tanθ }}{{secθ - tanθ }} = 5\) Concept used: secθ = Hypotenuse/Base, tanθ = perpendicular/base, cosθ = base/hypotenuse Pythgorus theorem Hypotenuse² = Perpendicular² + Base² Calculation: Let Perpendicular = P, Base = B, Hypotenuse = H \(\frac{{secθ + tanθ }}{{secθ - tanθ }} = 5\) ⇒ {(H/B) + (P/B)}/{(H/B) - (P/B)} = 5/1 ⇒ (H + P)/(H - P) = 5/1 ⇒ H + P = 5 ----(1) ⇒ H - P = 1 ----(2) Solve (1) and (2) ⇒ H = 3, P = 2 Hypotenuse2 = Perpendicular2 + Base2 ⇒ Base = √5 \(\frac{{3{{\cos }^2}\theta + 1}}{{3{{\cos }^2}\theta - 1}}\) ⇒ {3 × (√5/3)² + 1}/{3 × (√5/3)2 - 1} ⇒ (8/3)/(2/3) ⇒ 4 ∴ The value is 4.

If \(\frac{{sec\theta + tan\theta }}{{sec\theta - tan\theta }} = 5\) and θ is an acute angle, then the value of \(\frac{{3{{\cos }^2}\theta + 1}}{{3{{\cos }^2}\theta - 1}}\) is: 1. 42. 33. 14. 2

Answer» Correct Answer - Option 1 : 4

Given:

\(\frac{{secθ + tanθ }}{{secθ - tanθ }} = 5\)

Concept used:

secθ = Hypotenuse/Base, tanθ = perpendicular/base, cosθ = base/hypotenuse

Pythgorus theorem

Hypotenuse² = Perpendicular² + Base²

Calculation:

Let Perpendicular = P, Base = B, Hypotenuse = H

\(\frac{{secθ + tanθ }}{{secθ - tanθ }} = 5\)

⇒ {(H/B) + (P/B)}/{(H/B) - (P/B)} = 5/1

⇒ (H + P)/(H - P) = 5/1

⇒ H + P = 5 ----(1)

⇒ H - P = 1 ----(2)

Solve (1) and (2)

⇒ H = 3, P = 2

Hypotenuse2 = Perpendicular2 + Base2

⇒ Base = √5

\(\frac{{3{{\cos }^2}\theta + 1}}{{3{{\cos }^2}\theta - 1}}\)

⇒ {3 × (√5/3)² + 1}/{3 × (√5/3)2 - 1}

⇒ (8/3)/(2/3)

⇒ 4

∴ The value is 4.