If cot theta + tan theta = 3 and 1 - cos^2 th | vedclass

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(C) Given $\cot 	heta + 	an 	heta = 3$. Since $\cot 	heta + 	an 	heta = \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = \fra

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$9 \alpha^2(6 - \alpha^2) = 1$

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(C) Given $\cot 	heta + 	an 	heta = 3$. Since $\cot 	heta + 	an 	heta = \frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = \frac{\cos^2 	heta + \sin^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta}$,we have $\frac{1}{\sin 	heta \cos 	heta} = 3$,so $\sin 	heta \cos 	heta = \frac{1}{3}$. Also,$1 - \cos^2 	heta - \alpha \cos 	heta = 0$ implies $\sin^2 	heta = \alpha \cos 	heta$. Squaring both sides,$\sin^4 	heta = \alpha^2 \cos^2 	heta = \alpha^2(1 - \sin^2 	heta)$. From $\sin 	heta \cos 	heta = \frac{1}{3}$,we have $\cos 	heta = \frac{1}{3 \sin 	heta}$. Substituting into $\sin^2 	heta = \alpha \cos 	heta$,we get $\sin^2 	heta = \alpha \left(\frac{1}{3 \sin 	heta}ight)$,so $\sin^3 	heta = \frac{\alpha}{3}$. Thus,$\sin^2 	heta = \left(\frac{\alpha}{3}ight)^{2/3}$. Substituting this into $\sin^4 	heta = \alpha^2(1 - \sin^2 	heta)$,we get $\left(\frac{\alpha}{3}ight)^{4/3} = \alpha^2 \left(1 - \left(\frac{\alpha}{3}ight)^{2/3}ight)$. Dividing by $\alpha^2$,we have $\frac{\alpha^{4/3}}{3^{4/3} \alpha^2} = 1 - \left(\frac{\alpha}{3}ight)^{2/3}$,which simplifies to $\frac{1}{3^{4/3} \alpha^{2/3}} = 1 - \left(\frac{\alpha}{3}ight)^{2/3}$. Alternatively,from $\sin^2 	heta = \alpha \cos 	heta$,we have $\cos 	heta = \frac{\sin^2 	heta}{\alpha}$. Since $\sin^2 	heta \cos^2 	heta = \frac{1}{9}$,we have $\sin^2 	heta \left(\frac{\sin^4 	heta}{\alpha^2}ight) = \frac{1}{9}$,so $\sin^6 	heta = \frac{\alpha^2}{9}$. Since $\sin^3 	heta = \frac{\alpha}{3}$,this is consistent. Using $\cos^2 	heta = 1 - \sin^2 	heta = 1 - (\frac{\alpha}{3})^{2/3}$,and $\cos^2 	heta = \frac{\sin^4 	heta}{\alpha^2} = \frac{(\alpha/3)^{4/3}}{\alpha^2} = \frac{\alpha^{4/3}}{3^{4/3} \alpha^2} = \frac{1}{3^{4/3} \alpha^{2/3}}$. Equating these,$1 - (\frac{\alpha}{3})^{2/3} = \frac{1}{3^{4/3} \alpha^{2/3}}$. Multiplying by $3^{4/3} \alpha^{2/3}$,we get $3^{4/3} \alpha^{2/3} - 3^{4/3} \frac{\alpha^{4/3}}{3^{2/3}} = 1$,so $3 \cdot 3^{1/3} \alpha^{2/3} - 3 \alpha^{4/3} = 1$. Cubing both sides leads to $9 \alpha^2(6 - \alpha^2) = 1$.

If $\cot \theta + \tan \theta = 3$ and $1 - \cos^2 \theta - \alpha \cos \theta = 0$,then

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