JEE Advance - Mathematics (1989 - No. 23)
If $$x = \sec \theta - \cos \theta $$ and $$y = {\sec ^n}\theta - {\cos ^n}\theta $$, then show
that $$\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$$
that $$\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$$
The question asks to prove a relationship between x, y, and their derivatives with respect to θ.
The question requires knowledge of trigonometric identities and differentiation.
The question involves algebraic manipulation to arrive at the given equation.
The provided information does not align with standard multiple-choice question formats.
The question is about solving a system of linear equations.