Question

If secθ + tanθ = a; where 0° < θ < 90° then find the value of sinθ.
1. 2a
2. 2a/(a² - 1)
3. (a² - 1)/(a² + 1)
4. 2a/(a² + 1)

Answer 1

Correct Answer - Option 3 : (a² - 1)/(a² + 1)

Given:

secθ + tanθ = a

Concept used:

secθ = 1/cosθ 

cosθ = Base/Hypotenuse

tanθ = sinθ/cosθ 

sinθ = Perpendicular/Hypotenuse

tan²θ + 1 = sec²θ

sec2θ – tan2θ = 1

(secθ + tanθ)(secθ – tanθ) = 1

secθ – tanθ = 1/(sec θ + tan θ))

Calculation:

secθ + tanθ = a      ----(I)

⇒ sec θ - tan θ = 1/a      ----(II)

Adding (I) and (II);

2sec θ = a + (1/a)

⇒ 2/cos θ = (a² + 1)/a

⇒ 1/cosθ = (a2 + 1)/2a

⇒ cosθ = 2a/(a2 + 1)

Putting this value in a right angled triangle;

cosθ = Base/Hypotenuse

Using Pythagoras theorem;

(H)² = (B)² + (P)² ,

⇒ (a2 + 1)² = (2a)² + (P)² ,

⇒ (P)2 = (a2 + 1)2 - (2a)2 

⇒ (P)2 = a⁴ + 1 + 2a² - 4a²

⇒ (P)2 = a4 + 1 - 2a2 

⇒ (P)2 = (a2 - 1)2 

⇒ P = (a2 - 1)

sinθ = Perpendicular/Hypotenuse

⇒ (a2 - 1)/(a2 + 1)

∴ The value of sinθ is (a2 - 1)/(a2 + 1)