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

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Question

(B) Given equation: $(m + 2)\sin 	heta + (2m - 1)\cos 	heta = 2m + 1$. Divide by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$): $(m + 2)	an 	heta

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given equation: $(m + 2)\sin 	heta + (2m - 1)\cos 	heta = 2m + 1$. Divide by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$): $(m + 2)	an 	heta + (2m - 1) = (2m + 1)\sec 	heta$. Squaring both sides: $((m + 2)	an 	heta + (2m - 1))^2 = (2m + 1)^2(1 + 	an^2 	heta)$. Let $	an 	heta = t$. Then $(m + 2)^2 t^2 + 2(m + 2)(2m - 1)t + (2m - 1)^2 = (2m + 1)^2(1 + t^2)$. Expanding and simplifying: $(m^2 + 4m + 4)t^2 + (4m^2 + 6m - 4)t + (4m^2 - 4m + 1) = (4m^2 + 4m + 1)(1 + t^2)$. $(m^2 + 4m + 4)t^2 + (4m^2 + 6m - 4)t + (4m^2 - 4m + 1) = (4m^2 + 4m + 1) + (4m^2 + 4m + 1)t^2$. Rearranging terms: $(3m^2 - 3)t^2 - (4m^2 + 6m - 4)t + 4m = 0$. $(3t - 4)((m^2 - 1)t - 2m) = 0$. Thus,$t = \frac{4}{3}$ or $t = \frac{2m}{m^2 - 1}$. Therefore,$	an 	heta = \frac{4}{3}$ is a valid solution.

If $(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1$,then which of the following is true?

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