5.3 Given: \( \tan ^{2} x\left(\frac{1}{\tan ^{2} x}-1\right) \) 5.3.1 Prove that \( \tan ^{2} x\left(\frac{1}{\tan ^{2} x}-1\right)=\frac{\cos 2 x}{\cos ^{2} x} \)

Simplify the expression by following steps: - step0: Solution: \(\frac{\cos\left(2x\right)}{\cos\left(x^{2}\right)}\) - step1: Rewrite the expression: \(\cos^{-1}\left(x^{2}\right)\cos\left(2x\right)\) - step2: Rewrite the expression: \(\cos\left(2x\right)\cos^{-1}\left(x^{2}\right)\) - step3: Simplify: \(\cos\left(2x\right)\sec\left(x^{2}\right)\) - step4: Rewrite the expression: \(\sec\left(x^{2}\right)\cos\left(2x\right)\) Calculate or simplify the expression \( \tan^2(x)*(1/\tan^2(x)-1) \). Simplify the expression by following steps: - step0: Solution: \(\tan^{2}\left(x\right)\left(\frac{1}{\tan^{2}\left(x\right)}-1\right)\) - step1: Subtract the terms: \(\tan^{2}\left(x\right)\times \frac{1-\tan^{2}\left(x\right)}{\tan^{2}\left(x\right)}\) - step2: Reduce the fraction: \(1\times \left(1-\tan^{2}\left(x\right)\right)\) - step3: Multiply the terms: \(1-\tan^{2}\left(x\right)\) - step4: Calculate: \(1-\sec^{2}\left(x\right)+1\) - step5: Calculate: \(2-\sec^{2}\left(x\right)\) To prove that \[ \tan^{2} x\left(\frac{1}{\tan^{2} x}-1\right)=\frac{\cos 2x}{\cos^{2} x}, \] we will simplify both sides and show that they are equal. ### Step 1: Simplifying the Left Side Starting with the left side: \[ \tan^{2} x\left(\frac{1}{\tan^{2} x}-1\right). \] We can rewrite this as: \[ \tan^{2} x \left(\frac{1 - \tan^{2} x}{\tan^{2} x}\right) = \frac{\tan^{2} x (1 - \tan^{2} x)}{\tan^{2} x} = 1 - \tan^{2} x. \] Using the identity \( \tan^{2} x = \frac{\sin^{2} x}{\cos^{2} x} \), we can express \( 1 - \tan^{2} x \) as: \[ 1 - \frac{\sin^{2} x}{\cos^{2} x} = \frac{\cos^{2} x - \sin^{2} x}{\cos^{2} x} = \frac{\cos(2x)}{\cos^{2} x}. \] Thus, the left side simplifies to: \[ \frac{\cos(2x)}{\cos^{2} x}. \] ### Step 2: Simplifying the Right Side Now, let's simplify the right side: \[ \frac{\cos 2x}{\cos^{2} x}. \] This expression is already in its simplest form. ### Conclusion Since both sides simplify to the same expression: \[ \tan^{2} x\left(\frac{1}{\tan^{2} x}-1\right) = \frac{\cos 2x}{\cos^{2} x}, \] we have proven the statement.