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

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(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: $((m + 2)	an 	heta + (2m - 1))^2 = (2m + 1)^2 \sec^2 	heta$.Using $\sec^2 	heta = 1 + 	an^2 	heta$ and letting $t = 	an 	heta$:$(m + 2)^2 t^2 + 2(m + 2)(2m - 1)t + (2m - 1)^2 = (2m + 1)^2 (1 + t^2)$.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:$(m^2 + 4m + 4 - 4m^2 - 4m - 1)t^2 + (4m^2 + 6m - 4)t + (4m^2 - 4m + 1 - 4m^2 - 4m - 1) = 0$.$(-3m^2 + 3)t^2 + (4m^2 + 6m - 4)t - 8m = 0$.$3(1 - m^2)t^2 + (4m^2 + 6m - 4)t - 8m = 0$.Factoring the quadratic equation:$(3t - 4)((1 - m^2)t + 2m) = 0$.Thus,$t = \frac{4}{3}$ or $t = \frac{2m}{m^2 - 1}$.

$(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1$,then $\tan \theta$ is equal to:

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