If $\mathrm{P}= an 15^{\circ}+\cot 15^{\circ}, \mathrm{Q}= an 22 \frac{1^{\circ}}{2}+\cot 22 \frac{1}{2}_2^{\circ}$ and $\mathrm{R}=\sin 54^{\circ}+\sin 18^{\circ}$, then their ascending order is

Question

If $\mathrm{P}=	an 15^{\circ}+\cot 15^{\circ}, \mathrm{Q}=	an 22 \frac{1^{\circ}}{2}+\cot 22 \frac{1}{2}_2^{\circ}$ and $\mathrm{R}=\sin 54^{\circ}+\sin 18^{\circ}$, then their ascending order is

Aicharya · Accepted Answer

Given, $P=	an 15^{\circ}+\cot 15^{\circ}, Q=	an 22 \frac{1^{\circ}}{2}+\cot 22 \frac{1^{\circ}}{2}$ and $R=\sin 54^{\circ}+\sin 18^{\circ}$ Now $P=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}+\frac{\cos 15^{\circ}}{\sin 15^{\circ}}=\frac{\sin ^2 15^{\circ}+\cos ^2 15^{\circ}}{\cos 15^{\circ} \sin 15^{\circ}}$ $ \begin{aligned} & \Rightarrow \quad P=\frac{1}{\cos 15^{\circ} \sin 15^{\circ}} 	imes \frac{2}{2}=\frac{2}{\sin 2 	imes 15^{\circ}}=2 \operatorname{cosec} 30^{\circ} \ & \Rightarrow \quad P=2 	imes 2=4 \end{aligned} $ $ \begin{aligned} & 	ext { and } Q=	an 22 \frac{1^{\circ}}{2}+\cot 22 \frac{1^{\circ}}{2}=\frac{\sin ^2\left(22 \frac{1^{\circ}}{2}ight)+\cos ^2\left(22 \frac{1^{\circ}}{2}ight)}{\sin \left(22 \frac{1^{\circ}}{2}ight) \cos \left(22 \frac{1^{\circ}}{2}ight)} \ & \Rightarrow Q=\frac{2}{2} 	imes \frac{1}{\sin \left(22 \frac{1^{\circ}}{2}ight) \cos \left(22 \frac{1^{\circ}}{2}ight)}=\frac{2}{\sin 45^{\circ}} \ & \Rightarrow Q=2 \sqrt{2} \approx 2.83 \ & 	ext { and } R=\sin 54^{\circ}+\sin 18^{\circ}=\sin (90-36)+\sin 18^{\circ} \ & =\cos 36^{\circ}+\sin 18^{\circ} \end{aligned} $ we know that $ \begin{aligned} & \sin 18^{\circ}=\frac{\sqrt{5}-1}{4} 	ext { and } \cos 36^{\circ}=\frac{\sqrt{5}+1}{4} \ & 	herefore R=\frac{\sqrt{5}+1}{4}+\frac{\sqrt{5}-1}{4}=\frac{2 \sqrt{5}}{4}=\frac{\sqrt{5}}{2} \approx 1.19 \end{aligned} $ Clearly $R < Q < P \Rightarrow$ option (c) is correct

If $\mathrm{P}=\tan 15^{\circ}+\cot 15^{\circ}, \mathrm{Q}=\tan 22 \frac{1^{\circ}}{2}+\cot 22…

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