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

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(C) Given: $\sin 	heta + \cos 	heta = m$ and $\sec 	heta + 	ext{cosec} 	heta = n$. Squaring the first equation: $(\sin 	heta + \cos 	heta)^2 =

Vedclass · Accepted Answer

$2m$

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(C) Given: $\sin 	heta + \cos 	heta = m$ and $\sec 	heta + 	ext{cosec} 	heta = n$. Squaring the first equation: $(\sin 	heta + \cos 	heta)^2 = m^2 \implies 1 + 2 \sin 	heta \cos 	heta = m^2 \implies 2 \sin 	heta \cos 	heta = m^2 - 1$. Now,consider the expression $n(m + 1)(m - 1) = n(m^2 - 1)$. Substitute $n = \sec 	heta + 	ext{cosec} 	heta = \frac{1}{\cos 	heta} + \frac{1}{\sin 	heta} = \frac{\sin 	heta + \cos 	heta}{\sin 	heta \cos 	heta} = \frac{m}{\sin 	heta \cos 	heta}$. Thus,$n(m^2 - 1) = \left( \frac{m}{\sin 	heta \cos 	heta} ight) (2 \sin 	heta \cos 	heta) = 2m$.

If $\sin \theta + \cos \theta = m$ and $\sec \theta + \text{cosec} \theta = n$,then $n(m + 1)(m - 1) = $

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