Show the given trigonometric equation — task. Maths CBSE Live product, Class 10.

Show that

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

\(=\)

\frac{2}{2 \sin^{2} β - 1}

Proof:

LHS \(=\)

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

\(=\)

\frac{2}{2 \sin^{2} β - 1}

\(=\) RHS

Answer variants:

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

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

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

\frac{1 + 1}{{2 \sin}^{2} β - 1}

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

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

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

Did you find an error?

4. Show the given trigonometric equation