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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

cosec 48° + cosec 96° + cosec 192° + cosec 384° =

- 2
- 1
0
$\frac{\sqrt{3}}{2}$

Solution:

Let's consider

x = \csc 48^{\circ} + \csc 96^{\circ} + \csc 192^{\circ} + \csc 384^{\circ}

So

x = \csc(90^{\circ} - 42^{\circ}) + \csc 96^{\circ} + \csc(270^{\circ} - 78^{\circ}) + \csc\left(\begin{array}{c} 360^{\circ} \\ +24^{\circ} \end{array}\right)

= \sec 42^{\circ} + \csc 96^{\circ} - (\sec 78^{\circ}) + \csc 24^{\circ}

= \;\frac{1}{\cos 42^{\circ}} + \;\frac{1}{\cos 78^{\circ}} + \;\frac{1}{\sin 96^{\circ}} + \;\frac{1}{\sin 24^{\circ}}

= \;\frac{\cos 78^{\circ} - \cos 42^{\circ}}{\cos 42^{\circ} \times \cos 78^{\circ}} + \;\frac{\sin 24^{\circ} + \sin 96^{\circ}}{\sin 96^{\circ} \times \sin 24^{\circ}}

Further simplify the above,

x = \;\frac{-2 \sin 60^{\circ} \times \sin 18^{\circ}}{\cos^{2} 60^{\circ} - \sin^{2} 18^{\circ}} + \;\frac{2 \sin 60^{\circ} \times \cos 36^{\circ}}{\sin^{2} 60^{\circ} - \sin^{2} 36^{\circ}}

= \;\frac{(-2)\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{5}-1}{4}\right)}{\left(\;\frac{1}{2}\right)^{2} - \left(\;\frac{\sqrt{5}-1}{4}\right)^{2}} + \;\frac{2\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{5}+1}{4}\right)}{\left(\;\frac{\sqrt{3}}{2}\right)^{2} - \left(\;\frac{\sqrt{10-2\sqrt{5}}}{4}\right)^{2}}

= -2\sqrt{3} + 2\sqrt{3}

= 0

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