If P = tan 15^{circ} + cot 15^{circ},Q = tan | vedclass

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(C) Given,$P = 	an 15^{\circ} + \cot 15^{\circ}$,$Q = 	an 22 \frac{1}{2}^{\circ} + \cot 22 \frac{1}{2}^{\circ}$ and $R = \sin 54^{\circ} + \sin 18^{

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$R, Q, P$

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(C) Given,$P = 	an 15^{\circ} + \cot 15^{\circ}$,$Q = 	an 22 \frac{1}{2}^{\circ} + \cot 22 \frac{1}{2}^{\circ}$ and $R = \sin 54^{\circ} + \sin 18^{\circ}$.For $P$: $P = \frac{\sin 15^{\circ}}{\cos 15^{\circ}} + \frac{\cos 15^{\circ}}{\sin 15^{\circ}} = \frac{\sin^2 15^{\circ} + \cos^2 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}} = \frac{1}{\frac{1}{2} \sin 30^{\circ}} = \frac{2}{1/2} = 4$.For $Q$: $Q = \frac{\sin^2(22.5^{\circ}) + \cos^2(22.5^{\circ})}{\sin 22.5^{\circ} \cos 22.5^{\circ}} = \frac{1}{\frac{1}{2} \sin 45^{\circ}} = \frac{2}{1/\sqrt{2}} = 2\sqrt{2} \approx 2.828$.For $R$: $R = \sin 54^{\circ} + \sin 18^{\circ} = \cos 36^{\circ} + \sin 18^{\circ} = \frac{\sqrt{5}+1}{4} + \frac{\sqrt{5}-1}{4} = \frac{2\sqrt{5}}{4} = \frac{\sqrt{5}}{2} \approx 1.118$.Comparing the values: $R \approx 1.118$,$Q \approx 2.828$,$P = 4$.Thus,the ascending order is $R < Q < P$.

If $P = \tan 15^{\circ} + \cot 15^{\circ}$,$Q = \tan 22 \frac{1}{2}^{\circ} + \cot 22 \frac{1}{2}^{\circ}$ and $R = \sin 54^{\circ} + \sin 18^{\circ}$,then their ascending order is

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