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Doctors Medicines MCQs Q&A

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We use $sin2A=2sinAcosA$ and $cos2A=2cos^2A-1$

hence $sin((2pi)/7)=2sin(pi/7)cos(pi/7)$ and $cos((2pi)/7)=2cos^2(pi/7)-1$

Hence, $sin((2pi)/7)/(1+cos((2pi)/7))=(2sin(pi/7)cos(pi/7))/(1+2cos^2(pi/7)-1)$ or

= $(2sin(pi/7)cos(pi/7))/(2cos^2(pi/7))$ or

= $sin(pi/7)/(cos(pi/7))=tan(pi/7)$

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