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 show that $\frac{a^2}{b^2}=\frac{(d-a)(c-a)}{(b-c)(b-d)}$.

Answer 1

$a{\sin }^{2}x+b{\cos }^{2}x=c\Rightarrow a{\sin }^{2}x+b(1-\sin x)=c\Rightarrow a{\sin }^{2}x+b-b{\sin }^{2}x=c\Rightarrow (a-b){\sin }^{2}x=c-b$
$\Rightarrow {\sin }^{2}x=\frac{c-b}{a-b}a{\sin }^{2}x+b{\cos }^{2}x=c\Rightarrow a(1-{\cos }^{2}x)+b{\cos }^{2}x=c\Rightarrow a-a{\cos }^{2}x+b{\cos }^{2}x=c\Rightarrow (b-a){\cos }^{2}x=c-a$
$\Rightarrow {\cos }^{2}x=\frac{c-b}{a-b}$
$\therefore {\tan }^{2}x=\frac{c-b}{a-c}$
$b{\sin }^{2}y+a{\cos }^{2}y=d\Rightarrow b{\sin }^{2}y+a(1-{\sin }^{2}y)=d\Rightarrow b{\sin }^{2}y+a-a{\sin }^{2}y=d\Rightarrow (b-a){\sin }^{2}y=d-a$
$\Rightarrow {\sin }^{2}y=\frac{d-a}{b-a}$
$b{\sin }^{2}y+a{\cos }^{2}y=d\Rightarrow b(1-{\cos }^{2}y)+a{\cos }^{2}y=d\Rightarrow b-b{\cos }^{2}y+a{\cos }^{2}y=d\Rightarrow (a-d){\cos }^{2}y=d-b$
$\Rightarrow {\cos }^{2}y=\frac{d-b}{a-b}$
$\therefore {\tan }^{2}y=\frac{d-b}{a-d}$
$a\tan x=b\tan y\Rightarrow \frac{\tan x}{\tan y}=\frac{b}{a}\Rightarrow \frac{{\tan }^{2}x}{{\tan }^{2}y}=\frac{b^2}{a^2}$
$\Rightarrow \frac{a^2}{b^2}=\frac{(d-a)(c-a)}{(b-c)(b-d)}$