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