Question

$cosec \; 48^{\circ} + cosec \; 96^{\circ} + cosec \; 192^{\circ} + cosec \; 384^{\circ} $ =

• A. -2
• B. -1
• C. 0
• D. $\frac{ \sqrt{3}}{2}$

Answer 1

Answer: (C)

$cosec \, 48^{\circ}+cosec \, 96^{\circ}+ cosec \, 192^{\circ}+cosec \, 384^{\circ}$
$=cosec \left(90^{\circ}-42^{\circ}\right)+cosec\, 96^{\circ}+cosec\left(270^{\circ}-78^{\circ}\right)$
$+cosec\left(360^{\circ}+24^{\circ}\right)$
$=\sec\, 42^{\circ}+cosec \, 96^{\circ}+-\sec\, 78^{\circ}+cosec \, 24^{\circ}$
$\left[\because cosec\left(90^{\circ}-\theta\right)=\sec\, \theta, cosec\left(270^{\circ}-\theta\right)=-\sec \,\theta\right]$
$=\frac{1}{\cos \,42^{\circ}}+\frac{1}{\sin \,96^{\circ}}-\frac{1}{\cos\, 78^{\circ}}+\frac{1}{\sin \, 24^{\circ}}$
$=\frac{1}{\cos \, 42^{\circ}}-\frac{1}{\cos \, 78^{\circ}}+\frac{1}{\sin \, 96^{\circ}}+\frac{1}{\sin\, 24^{\circ}}$
$=\frac{\left(\cos\, 78^{\circ}-\cos\, 42^{\circ}\right)}{\cos\, 42^{\circ} \times \cos \, 78^{\circ}}+\frac{\left(\sin\, 24^{\circ}+\sin \, 96^{\circ}\right)}{\sin \, 96^{\circ} \times \sin\, 24^{\circ}}$
$=\frac{-2 \, \sin\, 60^{\circ} \times \sin\, 18^{\circ}}{\cos \left(60^{\circ}-18^{\circ}\right) \cos \left(60^{\circ}+18^{\circ}\right)}+\frac{2\, \sin\, 60^{\circ} \times \cos \, 36^{\circ}}{\sin \left(60^{\circ}+36^{\circ}\right) \sin \left(60^{\circ}-36^{\circ}\right)}$
$=\frac{-2 \, \sin \, 60^{\circ} \times \sin \, 18^{\circ}}{\cos ^{2}\left(60^{\circ}\right)-\sin ^{2}\left(18^{\circ}\right)}+\frac{2\, \sin\, 60^{\circ} \times \cos\, 36^{\circ}}{\sin ^{2} \, 60-\sin ^{2} \, 36^{\circ}}$
$= \frac{-2 \times \sqrt{3} / 2 \times\left(\frac{\sqrt{5}-1}{4}\right)}{(1 / 2)^{2}-\left(\frac{\sqrt{5}-1)}{4}\right)^{2}}+\frac{2 \times \sqrt{3} / 2 \times\left(\frac{\sqrt{5}+1}{4}\right)}{(\sqrt{3} / 2)^{2}-\left(\frac{\sqrt{10-2 \sqrt{5}}}{4}\right)^{2}} $
$=-2 \sqrt{3}+2 \sqrt{3}=0$