If x = asin θ, and y = btanθ , then find the value of $$\frac{a^2}{x^2} -\frac{b^2}{y^2 }$$ .​

Correct option is A

Given:
x = a sin θ
y = b tan θ
Formula Used:
From trigonometric identity; 1 + cot² θ = cosec²θ
Solution:
​$$=> \frac{a^2 }{ x^2}- \frac{b^2}{ y^2} = \frac{a^2 }{ (a sin θ) ^2} -\frac{ b^2 }{ (b tan θ)^2} \\\ \\=>\frac{ 1 }{sin^2 θ} -\frac{ 1 }{ tan^2θ }\\\ \\=> cosec^2 θ - cot^2 θ$$​​
From trigonometric identity 

​$$1 + cot^2 θ = cosec^2θ \\\ \\=> cosec^2 θ - cot^2 θ = 1$$​​