Question

$cosec48^{\circ }+cosec96^{\circ }+cosec192^{\circ }+cosec384^{\circ }$ =

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

Answer 1

Answer: (C)

$cosec48^{\circ }+cosec96^{\circ }+cosec192^{\circ }+cosec384^{\circ }$
$=cosec\left(90^{\circ }-42^{\circ }\right)+cosec96^{\circ }+cosec\left(270^{\circ }-78^{\circ }\right)$
$+cosec\left(360^{\circ }+24^{\circ }\right)$
$=\sec 42^{\circ }+cosec96^{\circ }+-\sec 78^{\circ }+cosec24^{\circ }$
$\left[\because cosec\left(90^{\circ }-\theta \right)=\sec \theta ,cosec\left(270^{\circ }-\theta \right)=-\sec \theta \right]$
$=\frac{1}{\cos 42^{\circ }}+\frac{1}{\sin 96^{\circ }}-\frac{1}{\cos 78^{\circ }}+\frac{1}{\sin 24^{\circ }}$
$=\frac{1}{\cos 42^{\circ }}-\frac{1}{\cos 78^{\circ }}+\frac{1}{\sin 96^{\circ }}+\frac{1}{\sin 24^{\circ }}$
$=\frac{\left(\cos 78^{\circ }-\cos 42^{\circ }\right)}{\cos 42^{\circ }\times \cos 78^{\circ }}+\frac{\left(\sin 24^{\circ }+\sin 96^{\circ }\right)}{\sin 96^{\circ }\times \sin 24^{\circ }}$
$=\frac{-2\sin 60^{\circ }\times \sin 18^{\circ }}{\cos \left(60^{\circ }-18^{\circ }\right)\cos \left(60^{\circ }+18^{\circ }\right)}+\frac{2\sin 60^{\circ }\times \cos 36^{\circ }}{\sin \left(60^{\circ }+36^{\circ }\right)\sin \left(60^{\circ }-36^{\circ }\right)}$
$=\frac{-2\sin 60^{\circ }\times \sin 18^{\circ }}{{\cos }^2\left(60^{\circ }\right)-{\sin }^2\left(18^{\circ }\right)}+\frac{2\sin 60^{\circ }\times \cos 36^{\circ }}{{\sin }^{2}60-{\sin }^{2}36^{\circ }}$
$=\frac{-2\times \sqrt{3}/2\times \left(\frac{\sqrt{5}-1}{4}\right)}{(1/2{)}^2-{\left(\frac{\sqrt{5}-1)}{4}\right)}^2}+\frac{2\times \sqrt{3}/2\times \left(\frac{\sqrt{5}+1}{4}\right)}{(\sqrt{3}/2{)}^2-{\left(\frac{\sqrt{10-2\sqrt{5}}}{4}\right)}^2}$
$=-2\sqrt{3}+2\sqrt{3}=0$