If secθ - tanθ = x/y, (0 < x < y) and 0ᵒ < θ < 90ᵒ, then sinθ is equal to:

If secθ - tanθ = x/y, (0 < x < y) and 0ᵒ < θ < 90ᵒ, then…

If secθ - tanθ = x/y, (0 < x < y) and 0^o < θ < 90^o, then sinθ is equal to:

Show answer & explanation

Correct answer: C

Explanation

The correct answer is Option 3 i.e (y² - x²)/(x² + y²)

Understanding/Application

secθ - tanθ = x/y ...(1

1/secθ - tanθ = y/x

(sec²θ - tan²θ)/secθ - tanθ = y/x

{(secθ - tanθ) (secθ + tanθ)}/secθ - tanθ = y/x

secθ + tanθ = y/x ....(2

Adding eq 1 and 2 -

secθ - tanθ + secθ + tanθ = x/y + y/x

2secθ = (x² + y²)/(xy)

secθ = (x² + y²)/2xy

cosθ = 2xy/(x² + y²)

b = 2xy

h = (x² + y²)

Using Pyth. theorem -

p² + b² = h²

p² + (2xy)² = (x² + y²)²

p² = x⁴ + y⁴ + 2x²y² - 4x²y²

p² = x⁴ + y⁴ - 2x²y²

p² = (x² - y²)²

p = x² - y² = y² - x²

sinθ = p/h = (y² - x²)/(x² + y²)