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

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Question

(c) $n({m^2} - 1) = (\sec 	heta + {m{cosec}}	heta ).2\sin 	heta \cos 	heta $                  &nbsp

Vedclass · Accepted Answer

$2m$

Vedclass · Answer

(c) $n({m^2} - 1) = (\sec 	heta + {m{cosec}}	heta ).2\sin 	heta \cos 	heta $                                                                                ($\because$ ${m^2} = 1 + 2\sin 	heta \cos 	heta $)

$ = \frac{{\sin 	heta + \cos 	heta }}{{\sin 	heta .\cos 	heta }}.2\sin 	heta \cos 	heta = 2m$.

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

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