If \( a \sin ^{2} x+b \cos ^{2} x=c, \quad b \sin ^{2} y+a \cos ^{2} y=d \) and \( a \tan x=b \tan y \), then \( \frac{a^{2}}{b^{2}} \) is equal to

To solve the problem, we start with the given equations: 1. \( a \sin^2 x + b \cos^2 x = c \) 2. \( b \sin^2 y + a \cos^2 y = d \) 3. \( a \tan x = b \tan y \) From the third equation, we can express \( \tan x \) and \( \tan y \) in terms of \( a \) and \( b \): \[ \tan x = \frac{b}{a} \tan y \] Next, we can express \( \sin^2 x \) and \( \cos^2 x \) in terms of \( \tan x \): \[ \sin^2 x = \frac{\tan^2 x}{1 + \tan^2 x} \quad \text{and} \quad \cos^2 x = \frac{1}{1 + \tan^2 x} \] Substituting \( \tan x = \frac{b}{a} \tan y \) into these expressions gives: \[ \sin^2 x = \frac{\left(\frac{b}{a} \tan y\right)^2}{1 + \left(\frac{b}{a} \tan y\right)^2} = \frac{\frac{b^2}{a^2} \tan^2 y}{1 + \frac{b^2}{a^2} \tan^2 y} = \frac{b^2 \tan^2 y}{a^2 + b^2 \tan^2 y} \] \[ \cos^2 x = \frac{1}{1 + \left(\frac{b}{a} \tan y\right)^2} = \frac{1}{1 + \frac{b^2}{a^2} \tan^2 y} = \frac{a^2}{a^2 + b^2 \tan^2 y} \] Now substituting these into the first equation: \[ a \sin^2 x + b \cos^2 x = a \left(\frac{b^2 \tan^2 y}{a^2 + b^2 \tan^2 y}\right) + b \left(\frac{a^2}{a^2 + b^2 \tan^2 y}\right) \] Combining these terms gives: \[ \frac{ab^2 \tan^2 y + ba^2}{a^2 + b^2 \tan^2 y} = c \] This simplifies to: \[ \frac{b(a^2 + b^2 \tan^2 y)}{a^2 + b^2 \tan^2 y} = c \] Thus, we have: \[ b(a^2 + b^2 \tan^2 y) = c(a^2 + b^2 \tan^2 y) \] Now, we can analyze the second equation: \[ b \sin^2 y + a \cos^2 y \] Using similar substitutions for \( \sin^2 y \) and \( \cos^2 y \): \[ \sin^2 y = \frac{\tan^2 y}{1 + \tan^2 y}, \quad \cos^2 y = \frac{1}{1 + \tan^2 y} \] Substituting these into the second equation gives: \[ b \left(\frac{\tan^2 y}{1 + \tan^2 y}\right) + a \left(\frac{1}{1 + \tan^2 y}\right) = d \] Combining these terms gives: \[ \frac{b \tan^2 y + a}{1 + \tan^2 y} = d \] This leads to: \[ b \tan^2 y + a = d(1 + \tan^2 y) \] Rearranging gives: \[ b \tan^2 y + a = d + d \tan^2 y \] Thus, we have: \[ (b - d) \tan^2 y = d - a \] Now, we can relate \( \frac{a^2}{b^2} \) using the equations we derived. From \( a \tan x = b \tan y \), we can express \( \tan y \) in terms of \( \tan x \): \[ \tan y = \frac{a}{b} \tan x \] Substituting this into the equations derived from \( c \) and \( d \) will yield a relationship between \( a \) and \( b \). After manipulating the equations,