Show the given trigonometric equation — task. Mathematics State Board, Class 10.

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Answer variants:

\frac{{(\sin x - \cos x)}^{2} + {(\sin x + \cos x)}^{2}}{\sin^{2} x - \cos^{2} x}

\frac{(\sin x - \cos x) (\sin x - \cos x)}{(\sin x + \cos x) (\sin x - \cos x)} + \frac{(\sin x + \cos x) (\sin x + \cos x)}{(\sin x - \cos x) (\sin x + \cos x)}

\frac{1 - 2 \sin x \cos x + 1 + 2 \sin x \cos x}{\sin^{2} x - 1 + \sin^{2} x}

\frac{{(\sin x + \cos x)}^{2} + {(\sin x + \cos x)}^{2}}{\sin^{2} x + \cos^{2} x}

\frac{\sin^{2} x + \cos^{2} x - 2 \sin x \cos x + \sin^{2} x + \cos^{2} x + 2 \sin x \cos x}{\sin^{2} x - (1 - \sin^{2} x)}

\frac{1 + 1}{{2 \sin}^{2} x - 1}

\frac{(\sin x - \cos x) (\sin x + \cos x)}{(\sin x + \cos x) (\sin x + \cos x)} + \frac{(\sin x + \cos x) (\sin x - \cos x)}{(\sin x - \cos x) (\sin x - \cos x)}

Show that

\frac{\sin x - \cos x}{\sin x + \cos x} + \frac{\sin x + \cos x}{\sin x - \cos x}

\(=\)

\frac{2}{2 \sin^{2} x - 1}

Proof:

LHS \(=\)

\frac{\sin x - \cos x}{\sin x + \cos x} + \frac{\sin x + \cos x}{\sin x - \cos x}

\(=\)

\frac{2}{2 \sin^{2} x - 1}

\(=\) RHS

Did you find an error?

12. Show the given trigonometric equation