Question

$\frac{\sin 2x}{\sin x}-\frac{\cos 2x}{\cos x}=\sec x$

Explanation:

Step1: Use double - angle formulas

We know that \(\sin2x = 2\sin x\cos x\) and \(\cos2x=\cos^{2}x-\sin^{2}x\).
Substitute these into the left - hand side of the equation:

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Step2: Simplify the first fraction

Simplify \(\frac{2\sin x\cos x}{\sin x}\), the \(\sin x\) terms cancel out, and we get \(2\cos x\).
The expression becomes \(2\cos x-\frac{\cos^{2}x-\sin^{2}x}{\cos x}\).

Step3: Get a common denominator

Rewrite \(2\cos x\) as \(\frac{2\cos^{2}x}{\cos x}\).
The expression is \(\frac{2\cos^{2}x-(\cos^{2}x - \sin^{2}x)}{\cos x}\).

Step4: Expand the numerator

Expand the numerator: \(2\cos^{2}x-\cos^{2}x+\sin^{2}x=\cos^{2}x+\sin^{2}x\).
Since \(\cos^{2}x+\sin^{2}x = 1\), the expression is \(\frac{1}{\cos x}\).

Step5: Use the reciprocal identity

We know that \(\frac{1}{\cos x}=\sec x\).

Answer:

The left - hand side \(\frac{\sin2x}{\sin x}-\frac{\cos2x}{\cos x}\) simplifies to \(\sec x\), so the identity \(\frac{\sin2x}{\sin x}-\frac{\cos2x}{\cos x}=\sec x\) is verified.