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EAMCET Engineering 2006 Solved Paper

Section: Mathematics

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Question : 108 of 160

Marks: +1, -0

If

x=\tan 15^{\circ}, y=\csc 75^{\circ}

z=4\sin 18^{\circ}

$x < y < z$
$y < z < x$
$z < x < y$
$x < z < y$

Solution:

\because \;\; x=\tan 15^{\circ}=\tan (45^{\circ}-30^{\circ})

\;=\;\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}

\; \text{ and }\; y=\csc 75^{\circ}

\;=\;\frac{1}{\sin (45^{\circ}+30^{\circ})}

\;=\;\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}

\; \text{ 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

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