If sin theta + cos theta = m and sec theta + | vedclass

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(C) Given that $\sin 	heta + \cos 	heta = m$ and $\sec 	heta + \csc 	heta = n$.We need to find the value of $n(m + 1)(m - 1)$,which is $n(m^2 - 1)

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$2m$

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(C) Given that $\sin 	heta + \cos 	heta = m$ and $\sec 	heta + \csc 	heta = n$.We need to find the value of $n(m + 1)(m - 1)$,which is $n(m^2 - 1)$.First,square the first equation: $m^2 = (\sin 	heta + \cos 	heta)^2 = \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = 1 + 2 \sin 	heta \cos 	heta$.Thus,$m^2 - 1 = 2 \sin 	heta \cos 	heta$.Now,substitute $n = \sec 	heta + \csc 	heta = \frac{1}{\cos 	heta} + \frac{1}{\sin 	heta} = \frac{\sin 	heta + \cos 	heta}{\sin 	heta \cos 	heta} = \frac{m}{\sin 	heta \cos 	heta}$.Therefore,$n(m^2 - 1) = \left( \frac{m}{\sin 	heta \cos 	heta} ight) (2 \sin 	heta \cos 	heta) = 2m$.

If $\sin \theta + \cos \theta = m$ and $\sec \theta + \csc \theta = n$,then $n(m + 1)(m - 1) = $

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