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Question: If \(x = \cos 10{^\circ}\cos 20{^\circ}\cos 40{^\circ},\)then the value of \(x\) is...

If x=cos⁡10∘cos⁡20∘cos⁡40∘,x = \cos 10{^\circ}\cos 20{^\circ}\cos 40{^\circ},then the value of xx is

A

14tan⁡10∘\frac{1}{4}\tan 10{^\circ}

B

18cot⁡10∘\frac{1}{8}\cot 10{^\circ}

C

18cosec10∘\frac{1}{8}\text{cosec}10{^\circ}

D

18sec⁡10∘\frac{1}{8}\sec 10{^\circ}

Answer

18cot⁡10∘\frac{1}{8}\cot 10{^\circ}

Explanation

Solution

x=cos⁡10ocos⁡20ocos⁡40ox = \cos{}10^{o}\cos{}20^{o}\cos{}40^{o}

=12sin⁡10o[2sin⁡10ocos⁡10ocos⁡20ocos⁡40o]= \frac{1}{2\sin 10^{o}}\lbrack 2\sin{}10^{o}\cos{}10^{o}\cos{}20^{o}\cos{}40^{o}\rbrack

=12.2sin⁡10o[2sin⁡20ocos⁡20ocos⁡40o]= \frac{1}{2.2\sin 10^{o}}\lbrack 2\sin{}20^{o}\cos{}20^{o}\cos{}40^{o}\rbrack

=12.4sin⁡10o[2sin⁡40ocos⁡40o)=18sin⁡10o(sin⁡80o)= \frac{1}{2.4\sin 10^{o}}\lbrack 2\sin 40^{o}\cos 40^{o}) = \frac{1}{8\sin 10^{o}}(\sin 80^{o})

=18sin⁡10ocos⁡10o=18cot⁡10o= \frac{1}{8\sin 10^{o}}\cos{}10^{o} = \frac{1}{8}\cot{}10^{o}.