If 540^{circ}

Question

(C) Given $540^{\circ} < 	heta < 630^{\circ}$,which implies $	heta$ is in the $3^{rd}$ quadrant. Since $	an 	heta = \frac{5}{12}$,we have $\sin

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$1$

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(C) Given $540^{\circ} Since $	an 	heta = \frac{5}{12}$,we have $\sin 	heta = -\frac{5}{13}$ and $\cos 	heta = -\frac{12}{13}$. For $\frac{	heta}{2}$,we have $\frac{540^{\circ}}{2} In the $4^{th}$ quadrant,$\cos \frac{	heta}{2} > 0$ and $\sin \frac{	heta}{2} Using $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1$,we get $-\frac{12}{13} = 2 \cos^2 \frac{	heta}{2} - 1$,so $2 \cos^2 \frac{	heta}{2} = \frac{1}{13}$,which gives $\cos \frac{	heta}{2} = \frac{1}{\sqrt{26}}$. Then $\sin \frac{	heta}{2} = -\sqrt{1 - \cos^2 \frac{	heta}{2}} = -\sqrt{1 - \frac{1}{26}} = -\frac{5}{\sqrt{26}}$. The denominator is $\sqrt{-(12 \sec 	heta + 5 \operatorname{cosec} 	heta)} = \sqrt{-(12 \cdot \frac{13}{-12} + 5 \cdot \frac{13}{-5})} = \sqrt{-(-13 - 13)} = \sqrt{26}$. Substituting these values,the expression becomes $\frac{\frac{1}{\sqrt{26}} - 5(-\frac{5}{\sqrt{26}})}{\sqrt{26}} = \frac{\frac{1+25}{\sqrt{26}}}{\sqrt{26}} = \frac{26}{26} = 1$.

If $540^{\circ} < \theta < 630^{\circ}$ and $\tan \theta = \frac{5}{12}$,then $\frac{\cos \frac{\theta}{2} - 5 \sin \frac{\theta}{2}}{\sqrt{-(12 \sec \theta + 5 \operatorname{cosec} \theta)}} = $

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