Show the given trigonometric equation — task. Mathematics CBSE, 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)}^{2} + {(\sin X + \cos X)}^{2}}{\sin^{2} X - \cos^{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)}

\frac{1 - 2 \sin X \cos X + 1 + 2 \sin X \cos X}{\sin^{2} X - 1 + \sin^{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{(\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?

11. Show the given trigonometric equation