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 $\frac{a^2}{b^2}$ equals to

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

Answer 1

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

dividing equation $asi{n}^{2}x+bco{s}^{2}x=c$ by $co{s}^{2}x$

$ata{n}^{2}x+b=cse{c}^{2}x\Rightarrow ata{n}^{2}x+b=c(1+ta{n}^{2}x)\Rightarrow ta{n}^{2}x(c-a)=b-c\Rightarrow ta{n}^{2}x=\frac{b-c}{c-a}$

Similarly , dividing equation $bsi{n}^{2}y+aco{s}^{2}y=d$ by $co{s}^{2}y$

$bta{n}^{2}y+a=dse{c}^{2}y\Rightarrow bta{n}^{2}y+a=d(1+ta{n}^{2}y)\Rightarrow ta{n}^{2}y(d-b)=a-d\Rightarrow ta{n}^{2}y=\frac{a-d}{d-b}$
Now, $atanx=btany\Rightarrow a^{2}ta{n}^{2}x=b^{2}ta{n}^{2}y\Rightarrow \frac{a^2}{b^2}=\frac{ta{n}^{2}y}{ta{n}^{2}x}\Rightarrow \frac{a^2}{b^2}=\frac{\frac{a-d}{d-b}}{\frac{b-c}{c-a}}=\frac{(a-d)(c-a)}{(b-c)(d-b)}$

Therefore, Answer is $\frac{(a-d)(c-a)}{(b-c)(d-b)}$