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Find the sum to n terms, whose nth term ...

Find the sum to n terms, whose nth term is `tan[alpha+(n-1)beta]tan (alpha+nbeta)`.

Text Solution

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`T_(n)=tan[alpha+(n-1)beta]tan (alpha+nbeta)`
`tanbeta=tan[(alpha+nbeta)-{alpha+(n-1)beta}]`
`tanbeta=(tan(alpha+nbeta)-tan(alpha+(n-1)beta))/(1+tan(alpha+nbeta)tan(alpha+(n-1)beta))`
`:.1+T_(n)=cotbeta[tan(alpha+nbeta)-tan(alpha+(n-1)beta)]`
`T_(n)=cotbeta[tan(alpha+nbeta)-tan(alpha+(n-1)beta)]-1`
For `n=1`
`T_(1)=cotbeta[tan(alpha+beta)-tanalpha]-1`
For `n=2`
`T_(2)=cotbeta[tan(alpha+2beta)-tan(alpha+beta)]-1`
For `n=3`,
`T_(3)=cotbeta[tan(alpha+3beta)-tan(alpha+2beta)]-1`
` " " vdots " " vdots " " vdots " "`
For `n=n`,
`T_(n)=cotbeta[tan(alpha+nbeta)-tan(alpha+(n-1)beta)]-1`
`S_(n)=T_(1)+T_(2)+T_(3)+"......."+T_(n)=cotbeta[tan(alpha+nbeta)-tanalpha]-n`
`=(sin(alpha+nbeta)/(cos(alpha+nbeta))+(sinalpha)/(cosalpha))/(tan beta)-n=(sin(alpha+nbeta-alpha)/(cosalpha cos(alpha+nbeta)))/(tan beta)-n`
`=((sinnbeta)/(cos(alpha+nbeta)cos alpha)-n tan beta)/(tan beta)`.
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