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Prove that :`sintheta/cos[3theta]+sin[3theta]/cos[9theta]+sin[9theta]/cos[27theta]=1/2[tan[27theta]-tan[theta]]`

Answer» `tan3theta=tantheta=(sin3theta)/cos3theta-sintheta/costheta`
`=(sin3thetatantheta-sinthetacos3theta)/(cos3theta*costheta)`
`=sin(3theta-theta)/(cos3theta*costheta)`
`=(2sintheta)/(cos3theta)`
`sintheta/cos3theta=1/2(tan3theta-tantheta)-(1)`
`(sin3theta)/(cos9theta)=1/2(tan9theta-tan3theta)-(2)`
`(sin9theta)/(cos27theta)=1/2(tan27theta-tan9theta)-(3)`
adding equation 1,2 and3
`sintheta/(cos3theta)+(sin3theta)/(cos9theta)+(sin9theta)/(cos27theta)=1/2(tan27theta-tantheta)`.


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