Question

Let $C$ be right angle of a triangle $ABC,$ then $\frac{{\sin }^{2}A}{{\sin }^{2}B}-\frac{{\cos }^{2}A}{{\cos }^{2}B}$ is equal to

• A. $\frac{a^2-b^2}{ab}$
• B. $\frac{a^4-b^4}{a^2{b}^2}$
• C. $\frac{a^4+b^4}{a^2{b}^2}$
• D. $\frac{a^2+b^2}{ab}$

Answer 1

Answer: (C)

Given, $\angle C={90}^o$
in $\Delta ABC$
$\because$ $\cos C=\frac{b^2+a^2-c^2}{2ab}$
$\because$ $\cos {90}^o=\frac{b^2+a^2-c^2}{2ab}=0$
$\Rightarrow$ $b^2+a^2-c^2=0\Rightarrow a^2+b^2=c^2$ ..(i)
Now, $\frac{{\sin }^{2}A}{{\sin }^{2}B}-\frac{{\cos }^{2}A}{{\cos }^{2}B}$
$=\frac{{(ak)}^2}{{(bk)}^2}-\frac{{\left(\frac{b^2+c^2-a^2}{2bc}\right)}^2}{{\left(\frac{a^2+c^2-b^2}{2ac}\right)}^2}$
$=\frac{a^2}{b^2}-\frac{{(b^2+c^2-a^2)}^2}{{(a^2+c^2-b^2)}^2}\times \frac{4a^2{c}^2}{4b^2{c}^2}$
$=\frac{a^2}{b^2}-\frac{a^2}{b^2}.{\left(\frac{b^2+c^2-a^2}{a^2+c^2-b^2}\right)}^2$
$=\frac{a^2}{b^2}-\frac{a^2}{b^2}{\left\{\frac{b^2+a^2+b^2-a^2}{a^2+a^2+b^2-b^2}\right\}}^2$ [from Eq. (i)]
$=\frac{a^2}{b^2}-\frac{a^2}{b^2}\left\{\frac{2b^2}{2a^2}\right\}=\frac{a^2}{b^2}\left\{1-\frac{b^4}{a^4}\right\}$
$=\frac{a^2}{b^2}\times \frac{a^4-b^4}{a^4}=\frac{(a^4-b^4)}{a^2{b}^2}$