Question

Answer needed urgently!!!​

Answer 1

Answer:

1/9

Step-by-step explanation:

3x = Cosecθ => x = 1/3 Cosecθ

3/x = Cotθ = 1/x = 1/3 Cotθ

x² - 1/x² = Cosec²θ/9 - Cot²θ/9

            = 1/9(Cosec²θ - Cot²θ)

             = 1/9  (∵ Cosec²θ - Cot²θ = 1)

Answer 2

Given:

3x = cosec y ——> 1

Squaring on both sides,

(3x)^2 = (cosec y)^2

9x^2 = cosec^2 y

3*3x^2 = cosec^2 y

3x^2 = (cosec^2 y)/3 ——> 2

3/x = cot y ——> 3

Squaring on both sides,

(3/x)^2 = (cot y)^2

9/(x)^2 = (cot y)^2

(3*3)/(x)^2 = (cot y)^2

3/(x)^2 = (cot^2 y)/3 ——> 4

Subtracting equation 4 from equation 3,

3x^2 - (3/(x)^2) = ((cosec^2 y)/3) - ((cot^2 y)/3)

3(x^2 - (1/(x)^2)) = (1/3)*((cosec^2 y) - (cot^2 y))

3(x^2 - (1/(x)^2)) = (1/3)*(1)

3(x^2 - (1/(x)^2)) = (1/3) ——> Answer