## Prove that, (sinA+sin3A)/(cosA+cos3A) = tan2A

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Sol. Now,

(sinA+sin3A)/(cosA+cos3A)

= (sinA+3sinA-4sin^3A)/(cosA+4cos^3A-3cosA)

= (4sinA-4sin^3A)/(4cos^3A-2cosA)

= [4sinA(1-sin^2A)]/[2cosA(2cos^2A-1)]

= (4sinAcos^2A)/(2cosAcos2A)

= (2sinAcosA)/cos2A

= sin2A/cos2A

= tan2A

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