Question

plz answer the question. ​

Answer 1

Given

• tanθ + sinθ = a _____(1)
• tanθ - sinθ = b ______(2)

To Prove

• a² - b² = 4√(ab)

Solution

tanθ + sinθ = a

tanθ - sinθ = b

Adding (1) and (2) we have

⇒tanθ + sinθ + tanθ - sinθ = a + b

⇒2tanθ = a + b

Substracting (2) from (1) we have

⇒(tanθ + sinθ) - (tanθ - sinθ) = a - b

⇒ tanθ + sinθ - tanθ + sinθ = a - b

⇒ 2sinθ = a - b

and

ab = (tanθ+ sinθ)(tanθ - sinθ)

⇒ ab = tan²θ - sin²θ

⇒ ab = sin²θ/cos²θ - sin²θ

⇒ ab = (sin²θ - sin²θ.cos²θ)/cos²θ

⇒ ab = sin²θ(1 - cos²θ)/cos²θ

⇒ab = tan²θ.sin²θ

⇒√(ab) = tanθ.sinθ

Now

(a - b)(a + b) = 2tanθ×2sinθ

⇒ a² - b² = 4tanθ.sinθ

⇒a² - b² = 4√(ab)

Proved

Answer 2

Solutions:

tan(theta) + sin(theta) = a

tan(theta) - sin(theta) = b

adding the equation (1) and (2).

tan(theta) + sin(theta) - tan(theta) - sin(theta) = a + b

2sin(theta) = a + b

substrating equation (2) and (1)

(tan(theta) + sin(theta)) - (tan(theta) - sin(theta)) = (a - b)

tan(theta) + sin(theta) - tan(theta) - sin(theta) = a - b

2sin(theta) = a + b

ab = (tan(theta) + sin(theta))(tan(theta) - sin(theta))

ab = tan²(theta) - sin²(theta)

ab = sin²(theta)/cos²(theta) - sin²(theta)

ab = sin²(theta) (1 - cos²(theta)/cos²(theta)

ab = tan²(theta) × sin²(theta)

√ab = tan(theta) × sin(theta)

(a + b)(a - b) = 2tan(theta) × 2sin(theta)

a² - b² = 4tan(theta)•sin(theta)

a² - b² = 4√ab

silentlover45.❤️