(m + 2)sin theta + (2m - 1)cos theta = 2m + 1 | vedclass

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Question

(b) Squaring the given relation and putting $	an 	heta = t,$

${(m + 2)^2}\,{t^2} + 2(m + 2)\,(2m - 1)t + {(2m - 1)^2} = {(2m + 1)^2}\,(1 + {t^2})

Vedclass · Accepted Answer

$	an 	heta = \frac{4}{3}$

Vedclass · Answer

(b) Squaring the given relation and putting $	an 	heta = t,$

${(m + 2)^2}\,{t^2} + 2(m + 2)\,(2m - 1)t + {(2m - 1)^2} = {(2m + 1)^2}\,(1 + {t^2})$

$ \Rightarrow \,3\,(1 - {m^2})\,{t^2} + (4{m^2} + 6m - 4)\,t - 8m = 0$

$ \Rightarrow \,(3t - 4)\,[(1 - {m^2})\,t + 2m] = 0$,

which is true, if $t = 	an 	heta = \frac{4}{3}$ or $	an 	heta = \frac{{2m}}{{{m^2} - 1}}$.

$(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1,$ if

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