Question

In any $∆ABC$, the simplified form of $\left(\frac{\cos 2A}{a^2}\right)-\left(\frac{\cos 2B}{b^2}\right)$ is:

• A. $a^2-b^2$
• B. $\frac{1}{a^2-b^2}$
• C. $\frac{1}{a^2}-\frac{1}{b^2}$
• D. $a^2+b^2$

Answer 1

The correct option is C

$\frac{1}{a^2}-\frac{1}{b^2}$

Explanation for the correct option:

Finding the value of the expression:

$\begin{array}{rcl}\left(\frac{\cos 2A}{a^2}\right)-\left(\frac{\cos 2B}{b^2}\right)& =& \frac{1-2{\sin }^{2}A}{a^2}-\frac{1-2{\sin }^{2}B}{b^2}\left[by\cos 2\theta =1-2{\sin }^2\theta \right]\\ & =& \frac{1}{a^2}-\frac{2{\sin }^{2}A}{a^2}-\frac{1}{b^2}+\frac{2{\sin }^{2}B}{b^2}\\ & =& \frac{1}{a^2}-\frac{1}{b^2}+2\left(\frac{{\sin }^{2}B}{b^2}-\frac{{\sin }^{2}A}{a^2}\right)\\ & =& \frac{1}{a^2}-\frac{1}{b^2}+2\left(0\right)\left[\mathrm{by}\mathrm{sine}\mathrm{rule}\frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}}\right]\\ & =& \frac{1}{a^2}-\frac{1}{b^2}\end{array}$

Hence, option C is correct.