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If $x=\tan 15^{\circ}, y=\operatorname{cosec} 75^{\circ}$ and $z=4 \sin 18^{\circ}$, then :
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Verified Answer
The correct answer is:
$x < y < z$
$\because \quad x=\tan 15^{\circ}=\tan\left(45^{\circ}-30^{\circ}\right)$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}$
$=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
and $y=\operatorname{cosec} 75^{\circ}$
$=\frac{1}{\sin \left(45^{\circ}+30^{\circ}\right)}$
$=\frac{1}{\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}}$
$=\frac{1}{\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{2 \sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{6}-\sqrt{2}$
and $\quad z=4 \sin 18^{\circ}=4\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{5}-1$
It is clear from above that
$(2-\sqrt{3}) < (\sqrt{6}-\sqrt{2}) < (\sqrt{5}-1)$
$\Rightarrow \quad x < y < z$
$=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{3-1}$
$=\frac{3+1-2 \sqrt{3}}{2}=2-\sqrt{3}$
and $y=\operatorname{cosec} 75^{\circ}$
$=\frac{1}{\sin \left(45^{\circ}+30^{\circ}\right)}$
$=\frac{1}{\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}}$
$=\frac{1}{\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}}=\frac{2 \sqrt{2}}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}$
$=\frac{2 \sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{6}-\sqrt{2}$
and $\quad z=4 \sin 18^{\circ}=4\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{5}-1$
It is clear from above that
$(2-\sqrt{3}) < (\sqrt{6}-\sqrt{2}) < (\sqrt{5}-1)$
$\Rightarrow \quad x < y < z$
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