Question

If $x=\tan 15^{\circ },y=\mathrm{cosec}75^{\circ }$ 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
$(2-\sqrt{3})<(\sqrt{6}-\sqrt{2})<(\sqrt{5}-1)$
$\Rightarrow x<y<z$