1.

Prove that :`sintheta/(cos(3theta)) + (sin3theta ) /(cos9theta) + (sin 9theta) /(cos27theta) =1/2( tan27theta-tantheta)`

Answer» `tan3theta - tan theta =( sin3theta)/(cos3 theta) - sin theta/cos theta`
`= (sin3thetacostheta - sinthetacos3theta)/(cos3thetacos theta)`
`=sin(3theta-theta)/(cos3thetacos theta)`
`=sin(2theta)/(cos3thetacos theta)`
`=(2sinthetacostheta)/(cos3thetacos theta) = (2sintheta)/(cos3theta)`
`:. 1/2(tan3theta - tan theta) = (sintheta)/(cos3theta)`
Similarly,
`1/2(tan9theta - tan 3theta) = (sin3theta)/(cos9theta)`
Similarly,
`1/2(tan27theta - tan 9theta) = (sin9theta)/(cos27theta)`
`:. (sintheta)/(cos3theta)+ (sin3theta)/(cos9theta) +(sin3theta)/(cos9theta) = 1/2(tan3theta - tan theta+tan9theta - tan 3theta+tan27theta - tan 9theta)`
`=> (sintheta)/(cos3theta)+ (sin3theta)/(cos9theta) +(sin3theta)/(cos9theta) = 1/2(tan27 theta - tantheta)`


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