\begin{tabular}{|l|l|}\hline MATHEMATICS (PAPER 2) \\ \hline 5.3 & Given: \( \tan ^{2} x\left(\frac{1}{\tan ^{2} x}-1\right) \) \\ \hline 5.3 .1 & Prove that \( \tan ^{2} x\left(\frac{1}{\tan ^{2} x}-1\right)=\frac{\cos 2 x}{\cos ^{2} x} \) \\ \hline\end{tabular}

To embark on this fun mathematical journey, let’s dive into the trigonometric world of tangent and cosine! First, recall that \( \tan x = \frac{\sin x}{\cos x} \). Hence, \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \). Now, rewriting the expression, we have: \[ \tan^{2} x\left(\frac{1}{\tan^{2} x}-1\right) = \tan^{2} x \left(\frac{\cos^{2} x}{\sin^{2} x} - 1\right) = \tan^{2} x \left(\frac{\cos^{2} x - \sin^{2} x}{\sin^{2} x}\right) \] Since \( \cos^2 x - \sin^2 x = \cos 2x \), we substitute that in: \[ \tan^{2} x\left(\frac{\cos^{2} x - \sin^{2} x}{\sin^{2} x}\right) = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{\cos 2x}{\sin^2 x} = \frac{\cos 2x}{\cos^2 x} \] And there you have it, a proven equality with a twist of trigonometry! Now, what’s next? You might wonder why this trigonometric identity is relevant. Well, these identities help simplify complex expressions in calculus and physics. Whether you're calculating forces, analyzing oscillations, or even in wave mechanics, understanding these relationships allows you to navigate through the mathematical landscape with ease! So, grab a pencil, keep practicing these identities, and pretty soon, you’ll be solving the trickiest problems like a pro!