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 $a^2/b^2$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. $\frac{(b–c)(b–d)}{(a–c)(a–d)}$

Answer 1

The correct option is B

$\frac{(a–d)(c–a)}{(b–c)(d–b)}$

Step1. Expressing the given equations:

Given, $a{\sin }^{2}x+b{\cos }^{2}x=c$

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

$$\begin{gathered} a{\tan }^{2}x+b=cse{c}^{2}x \\ \Rightarrow a{\tan }^{2}x+b=c(1+{\tan }^{2}x) \\ \Rightarrow b–c=\left(\right(c–a){\tan }^{2}x \\ \Rightarrow {\tan }^{2}x=\frac{(b–c)}{(c–a)}.....\dots .(i) \end{gathered}$$

Also, given $b{\sin }^{2}y+a{\cos }^{2}y=d$

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

$$\begin{gathered} b{\tan }^{2}y+a=dse{c}^{2}y \\ \Rightarrow b\tan 2y+a=d(1+{\tan }^{2}y) \\ \Rightarrow a–d=(d–b){\tan }^{2}y \\ \Rightarrow {\tan }^{2}y=\frac{(a–d)}{(d–b)}......\dots .\left(ii\right) \end{gathered}$$

Step2. Find the value of ${\mathit{a}}^{\mathbf{2}}\mathbf{/}{\mathit{b}}^{\mathbf{2}}$:

Now, $a\tan x=b\tan y$

Squaring on both sides,

$$\begin{gathered} a^2{\tan }^{2}x=b^2{\tan }^{2}y \\ \Rightarrow \frac{a^2}{b^2}=\frac{{\tan }^{2}x}{{\tan }^{2}y} \end{gathered}$$

Substituting (i) and (ii),

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

Hence, the correct option is (B).