If `asin^2x+bcos^2x=c,bsin^2y+acos^2y=d` and `atanx=btany` then `a^2/b^2=........ (0ltx,yltpi/ 2)`

Options

A. `((b-c)(d-b))/((a-d)(d-b))`
B. `((a-d)(c-a))/((b-c)(d-b))`
C. `((d-a)(c-a))/((b-c)(d-b))`
D. `((b-c)(b-d))/((a-c)(a-d))`

Answer

To solve the problem, we need to find the value of \( \frac{a^2}{b^2} \) given the equations:

1. \( a \sin^2 x + b \cos^2 x = c \)
2. \( b \sin^2 y + a \cos^2 y = d \)
3. \( a \tan x = b \tan y \)

### Step 1: Express \( \frac{a}{b} \) in terms of \( \tan x \) and \( \tan y \)

From the third equation \( a \tan x = b \tan y \), we can rearrange it to get:

\[ \frac{a}{b} = \frac{\tan y}{\tan x} \]

### Step 2: Square both sides to find \( \frac{a^2}{b^2} \)

Squaring both sides gives us:

\[ \frac{a^2}{b^2} = \frac{\tan^2 y}{\tan^2 x} \]

### Step 3: Find \( \tan^2 x \)

Using the first equation \( a \sin^2 x + b \cos^2 x = c \), we can divide the entire equation by \( \cos^2 x \):

\[ a \tan^2 x + b = c \sec^2 x \]

Since \( \sec^2 x = 1 + \tan^2 x \), we can rewrite this as:

\[ a \tan^2 x + b = c + c \tan^2 x \]

Rearranging gives:

\[ a \tan^2 x - c \tan^2 x = c - b \]

Factoring out \( \tan^2 x \):

\[ \tan^2 x (a - c) = c - b \]

Thus, we find:

\[ \tan^2 x = \frac{c - b}{a - c} \]

### Step 4: Find \( \tan^2 y \)

Using the second equation \( b \sin^2 y + a \cos^2 y = d \), we divide by \( \cos^2 y \):

\[ b \tan^2 y + a = d \sec^2 y \]

Again, using \( \sec^2 y = 1 + \tan^2 y \):

\[ b \tan^2 y + a = d + d \tan^2 y \]

Rearranging gives:

\[ b \tan^2 y - d \tan^2 y = d - a \]

Factoring out \( \tan^2 y \):

\[ \tan^2 y (b - d) = d - a \]

Thus, we find:

\[ \tan^2 y = \frac{d - a}{b - d} \]

### Step 5: Substitute \( \tan^2 x \) and \( \tan^2 y \) into \( \frac{a^2}{b^2} \)

Now substituting these values into \( \frac{a^2}{b^2} \):

\[ \frac{a^2}{b^2} = \frac{\tan^2 y}{\tan^2 x} = \frac{\frac{d - a}{b - d}}{\frac{c - b}{a - c}} \]

This simplifies to:

\[ \frac{a^2}{b^2} = \frac{(d - a)(a - c)}{(b - d)(c - b)} \]

### Final Result

Thus, the value of \( \frac{a^2}{b^2} \) is:

\[ \frac{a^2}{b^2} = \frac{(d - a)(a - c)}{(b - d)(c - b)} \]