Question

If $asi{n}^{2}x+bco{s}^{2}x=c,$ $bsi{n}^{2}y+aco{s}^{2}y=d,$ and $atanx=btany,$ then $\frac{a^2}{b^2}$ is equal to

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

Answer 1

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

$asi{n}^{2}x+bco{s}^{2}x=c\Rightarrow (b-a)co{s}^{2}x=c-a$
$\Rightarrow (b-a)=(c-a)(1+ta{n}^{2}x)$
$bsi{n}^{2}y+aco{s}^{2}y=d\Rightarrow (a-b)co{s}^{2}y=d-b$
$\Rightarrow (a-b)=(d-b)(1+ta{n}^{2}y)$
$\therefore ta{n}^{2}x=\frac{b-c}{c-a},ta{n}^{2}y=\frac{a-d}{d-b}$
$\therefore \frac{ta{n}^{2}x}{ta{n}^{2}y}=\frac{(b-c)(d-b)}{(c-a)(a-d)}$ ....(i)
But $atanx=btany,i.e,\frac{tanx}{tany}=\frac{b}{a}$ ....(ii)
From (i) and (ii), $\frac{b^2}{a^2}=\frac{(b-c)(d-b)}{(c-a)(a-d)}$
$\Rightarrow \frac{a^2}{b^2}=\frac{(c-a)(a-d)}{(b-c)(d-b)}.$