Question

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

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

Answer 1

The correct option is B $\frac{(a-d)(c-a)}{(b-c)(d-b)}$

We have $a\tan x=b\tan y$

Squaring both sides, we get

$\Rightarrow a^2{\tan }^{2}x=b^2{\tan }^{2}y$

$\Rightarrow \frac{a^2}{b^2}=\frac{{\tan }^{2}y}{{\tan }^{2}x}$ .........$\left(1\right)$

We have $a{\sin }^{2}x+b{\cos }^{2}x=c$

Dividing both sides by ${\cos }^{2}x$ we get

$\Rightarrow a\frac{{\sin }^{2}x}{{\cos }^{2}x}+b=\frac{c}{{\cos }^{2}x}$

$\Rightarrow a{\tan }^{2}x+b=c{\sec }^{2}x$

$\Rightarrow a{\tan }^{2}x+b=c(1+{\tan }^{2}x)$

$\Rightarrow a{\tan }^{2}x+b=c+c{\tan }^{2}x$

$\Rightarrow a{\tan }^{2}x-c{\tan }^{2}x=c-b$

$\Rightarrow (a-c){\tan }^{2}x=c-b$

$\Rightarrow {\tan }^{2}x=\frac{c-b}{a-c}$ .........$\left(2\right)$

We have $b{\sin }^{2}y+a{\cos }^{2}y=d$

Dividing both sides by ${\cos }^{2}y$ we get

$\Rightarrow b\frac{{\sin }^{2}y}{{\cos }^{2}y}+a=\frac{d}{{\cos }^{2}y}$

$\Rightarrow b{\tan }^{2}y+a=d{\sec }^{2}y$

$\Rightarrow b{\tan }^{2}y+a=d(1+{\tan }^{2}y)$

$\Rightarrow b{\tan }^{2}y+a=d+d{\tan }^{2}y$

$\Rightarrow b{\tan }^{2}y-d{\tan }^{2}y=d-a$

$\Rightarrow (b-d){\tan }^{2}y=d-a$

$\Rightarrow {\tan }^{2}y=\frac{d-a}{b-d}$ .........$\left(3\right)$

Substituting $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$ we get

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

$\therefore \frac{a^2}{b^2}=\frac{(d-a)(a-c)}{(b-d)(c-b)}=\frac{(a-d)(c-a)}{(d-b)(b-c)}$