Question

If $x=\tan 15^{\circ}, y=\operatorname{cosec} 75^{\circ}$ and $z=4 \sin 18^{\circ}$, then :

• A. x < y < z
• B. y < z < x
• C. z < x < y
• D. x < z < y

Answer 1

Answer: (A)

$\because 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 =\text{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 $z=4 \sin 18^{\circ}=4\left(\frac{\sqrt{5}-1}{4}\right)=\sqrt{5}-1$
It is clear from above that
$(2-\sqrt{3})<(\sqrt{6}-\sqrt{2})<(\sqrt{5}-1)$
$\Rightarrow x < y < z$