Question

If $asin^{2} x+bcos^{2} ⁡ x=c, \, bsin^{2} ⁡ y+acos^{2} ⁡ y=d$ and $atan x=btan ⁡ y,$ then $\frac{a^{2}}{b^{2}}$ is equal to (Here $a, \, b, \, c$ and $d \, $ are distinct)

• A. $\frac{\left(b - c\right) \left(d - b\right)}{\left(a - d\right) \left(a - b\right)}$
• B. $\frac{\left(a - d\right) \left(c - a\right)}{\left(b - c\right) \left(d - b\right)}$
• C. $\frac{\left(d - a\right) \left(c - a\right)}{\left(b - c\right) \left(d - b\right)}$
• D. $\frac{\left(b - c\right) \left(b - d\right)}{\left(a - c\right) \left(a - d\right)}$

Answer 1

Answer: (B)

Dividing the given equations by $cos^{2} x \, \& \, cos^{2}⁡y$ respectively. We get,
$tan^{2}x=\frac{c - b}{a - c}$ and $tan^{2}y=\frac{d - a}{b - d}$
$\Rightarrow \frac{a^{2}}{b^{2}}=\frac{\left(tan\right)^{2} y}{\left(tan\right)^{2} x}=\frac{\left(\right. a - d \left.\right) \, \left(\right. c - a \left.\right)}{\left(\right. b - c \left.\right) \, \left(\right. d - b \left.\right)}$