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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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?

Similar Questions

If $\cos(x) + \cos^2(x) = 1$,then $\sin^2(x) + \sin^4(x)$ is equal to

$\tan ^2 \frac{\pi}{16}+\tan ^2 \frac{2 \pi}{16}+\tan ^2 \frac{3 \pi}{16}+\tan ^2 \frac{4 \pi}{16}+\tan ^2 \frac{5 \pi}{16}+\tan ^2 \frac{6 \pi}{16}+\tan ^2 \frac{7 \pi}{16} = ?$

$\operatorname{cosec} 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ}=$

If $\tan A + \cot A = 2$,then the value of $\tan^{4} A + \cot^{4} A$ is:

If $A$ and $B$ are positive acute angles satisfying $3 \cos^2 A + 2 \cos^2 B = 4$ and $\frac{3 \sin A}{\sin B} = \frac{2 \cos B}{\cos A}$,then $A + 2B =$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module